The tension in each brace and the reaction at C .

Question

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Answer 1

Question

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

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Answer 2

Question

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

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Answer 3

Question

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

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Answer 4

Question

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

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Answer 5

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Answer 6

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$

Resolve i, j, and k components as shown below.

Resolve j component.

$−1.8368TBD+1.8368TBE=01.8368TBD=1.8368TBETBD=TBE$ (1)

Resolve i component.

$−1.8369TBD−1.8369TBE−649.5=0$

Substitute $TBE$ for $TAD$ .

$−1.8369TBE−1.8369TBE−649.5=0−3.6738TBE=649.5TBE=−176.79 lbTBE=176.79 lb$

Hence, the tension in brace BE is $TBE=176.8 lb_$ .

Calculate the tension in brace BD as shown below.

Substitute $176.8 lb$ for $TBE$ in Equation (1).

$TBD=176.8 lb$

Hence, the tension in brace BD is $TBD=176.8 lb_$ .

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of forces in x-direction and equate to zero for equilibrium.

$∑Fx=0Cx+176.8×(−0.707)+176.8×0.707=0Cx=0$

Apply the equilibrium equation of forces in y-direction and equate to zero for equilibrium.

$∑Fy=0Cy+176.8×0.3535+176.8×0.3535−75=0Cy+50=0Cy=−50 lb$

Apply the equilibrium equation of forces in z-direction and equate to zero for equilibrium.

$∑Fz=0Cz+176.8×(−0.6123)+176.8×(−0.6123)=0Cz−216.5=0Cz=216.5 lb$

Calculate the reaction at C as shown below.

$C=Cxi+Cyj+Czk$

Substitute 0 for $Cx$ , $−50 lb$ for $Cy$ , and $216.5 lb$ for $Cz$ .

$C=(0)i+(−50)j+(216.5)k=−(50 lb)j+(216.5 lb)k$

Therefore, the reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Answer 7

Chapter 4.3, Problem 4.110P

To determine

The tension in each brace and the reaction at C.

Answer 8

Expert Solution & Answer

Answer to Problem 4.110P

The tension in brace BD is $TBD=176.8 lb_$ .

The tension in brace BE is $TBE=176.8 lb_$ .

The reaction at C is $C=−(50 lb)j+(216.5 lb)k_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

The length of the flag pole AC is $10 ft$ .

The inclination of the flag pole AC is $30°$ .

The distance BC is $3 ft$ .

Calculation:

Assumption:

Apply the sign convention for calculating the equations of equilibrium as shown below.

For the horizontal forces equilibrium condition, take the force acting towards right side as positive $(→+)$ and the force acting towards left side as negative $(←−)$ .
For the vertical forces equilibrium condition, take the upward force as positive $(↑+)$ and the downward force as negative $(↓−)$ .
For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive.

Draw the free body diagram as in Figure (1).

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 4.3, Problem 4.110P

Calculate the position vector $(r)$ as shown below.

The position of A is;

$rA=(10sin30°)j−(10cos30°)k=(5 ft)j−(8.66 ft)k$

The position of B is;

$rB=(3sin30°)j+(3cos30°)k=(1.5 ft)j+(2.598 ft)k$

The position of D is;

$rD=−(3 ft)i+(3 ft)j$

The position of E is;

$rE=(3 ft)i+(3 ft)j$

Calculate the position of brace BD as shown below.

$BD→=rD−rB$

Substitute $−(3 ft)i+(3 ft)j$ for $rD$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BD→=[−(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=−(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BD as shown below.

$BD=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the position of brace BE as shown below.

$BE=rE−rB$

Substitute $(3 ft)i+(3 ft)j$ for $rB$ and $(1.5 ft)j+(2.598 ft)k$ for $rB$ .

$BE→=[(3 ft)i+(3 ft)j]−[(1.5 ft)j+(2.598 ft)k]=(3 ft)i+(1.5 ft)j−(2.598 ft)k$

Calculate the length of brace BE as shown below.

$BE=(−3)2+(1.5)2+(−2.598)2=17.999604=4.243 ft$

Calculate the tension in brace BD $(TBD)$ as shown below.

$TBD=TBDBD→BD$

Substitute $−(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BD→$ and $4.243 ft$ for $BD$ .

$TBD=TBD−3i+1.5j−2.598k4.243=TBD(−0.707i+0.3535j−0.6123k)$

Calculate the tension in brace BE $(TBE)$ as shown below.

$TBE=TBEBE→BE$

Substitute $(3 ft)i+(1.5 ft)j−(2.598 ft)k$ for $BE→$ and $4.243 ft$ for $BE$ .

$TBE=TBE3i+1.5j−2.598k4.243=TBE(0.707i+0.3535j−0.6123k)$

Apply the Equations of Equilibrium as shown below.

Apply the equilibrium equation of moment about point C and equate to zero for equilibrium.

$∑MC=0rB×TBD+rB×TBE+rA×(−75j)=0$

Substitute $(5 ft)j−(8.66 ft)k$ for $rA$ , $(1.5 ft)j+(2.598 ft)k$ for $rB$ , $TBD(−0.707i+0.3535j−0.6123k)$ for $TBD$ , and $TBE(0.707i+0.3535j−0.6123k)$ for $TBE$ .

$[(1.5j+2.598k)×(TBD(−0.707i+0.3535j−0.6123k))+(1.5j+2.598k)×(TBE(0.707i+0.3535j−0.6123k))+(5j−8.66k)×(−75j)]=0TBD|ijk01.52.598−0.7070.3535−0.6123|+TBE|ijk01.52.5980.7070.3535−0.6123|+|ijk05−8.660−750|=0[TBD(i(−0.9185−0.9184)−j(0+1.8368)+k(0+1.0605))+TBE(i(−0.9185−0.9184)−j(0−1.8368)+k(0−1.0605))+(i(0−649.5)−j(0)+k(0))]=0(TBD(−1.8369i−1.8368j+1.0605k)+TBE(−1.8369i+1.8368j−1.0605k)−649.5i)=0$

Multiply by $4.243$ .

$(−1.8369TBDi−1.8368TBDj+1.0605TBDk−1.8369TBEi+1.8368TBEj−1.0605TBEk−649.5i)=0{(−1.8369TBD−1.8369TBE−649.5)i+(−1.8368TBD+1.8368TBE)+(1.0605TBD−1.0605TBEk)}=0$