Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 17.1, Problem 17.14P

The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder A and block B are attached to cords that are wrapped on the pulleys as shown. The coefficient of kinetic friction between block B and the surface is 0.2. Knowing that the system is at rest in the position shown when a constant force P = 200 N is applied to cylinder A, determine (a) the velocity of cylinder A as it strikes the ground, (b) the total distance that block B moves before coming to rest.

Chapter 17.1, Problem 17.14P, The double pulley shown has a mass of 15 kg and a centroidal radius of gyration of 160 mm. Cylinder

Fig. P17.14

(a)

Expert Solution
Check Mark
To determine

Find the velocity of cylinder A as it strikes the ground.

Answer to Problem 17.14P

The velocity of the cylinder A when it strikes the ground is 4.79m/s_.

Explanation of Solution

Given information:

The mass of the cylinder A is mA=5kg.

The mass of the block B is mB=15kg.

The radius of the outer pulley is rA=250mm.

The radius of the inner pulley is rB=150mm.

The centroidal radius of gyration is kC=160mm.

The coefficient of friction between the surface and the block B is μk=0.20.

The constant force applied at cylinder A is P=200N.

Calculation:

Consider the acceleration due to gravity is g=9.81m/s2.

Consider the radius of the outer pulley as rA.

Consider the radius of the inner pulley as rB.

Find the velocity in the outer pulley (vA) using the relation.

vA=rAωCωC=vArA (1)

Here, the angular velocity of the pulley is ωC.

Find the velocity in the inner pulley (vB) using the relation.

vB=rBωC

Substitute vArA for ωC.

vB=rBvArA (2)

Find the distance of the outer pulley (sA) is in contact using the relation.

sA=rAθCθC=sArA

Here, the number of revolutions in the pulley C is θC.

Find the distance of the inner pulley (sB) is in contact using the relation.

sB=rBθC

Substitute sArA for θC.

sB=rBsArA (3)

The initial total kinetic energy at rest is zero.

T1=0

Find the mass moment of inertia in the pulley C (I¯C) using the equation.

I¯C=mCkC2

Here, the mass in the pulley C is mC.

Substitute 15 kg for mC and 160 mm for kC.

I¯C=15×(160mm×1m1,000mm)2=0.384kgm2

Find the total kinetic energy (T2) at the final stage using the relation.

T2=12mAvA2+12mBvB2+12I¯CωC2

Substitute 5 kg for mA, 15 kg for mB, rBvArA for vB, 0.384kgm2 for I¯C, and vArA for ωC.

T2=12×5×vA2+12×15×(rBvArA)2+12×0.384×(vArA)2

Substitute 250 mm for rA and 150 mm for rB.

T2=12×5×vA2+12×15×(150250vA)2+12×0.384×(vA250mm×1m1,000mm)2=2.5vA2+2.7vA2+3.072vA2=8.272vA2

When sA=1m;

Substitute 150 mm for rB, 1 m for sA and 250 mm for rA in Equation (3).

sB=150×1250=0.6m

Show free-body diagram the block B as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 17.1, Problem 17.14P

Resolve the vertical component of forces as follows;

Fy=0NBmBgcos30°=0NB=15×9.81×cos30°=127.44N

Find the frictional force Ff using the relation.

Ff=μkNB

Substitute 0.20 for μk and 127.44 N for NB.

Ff=0.20×127.44=25.487N

Apply the principle of work and energy for the cylinder A, the block B and the double pulley C as follows;

U12=PsA+mAgsAFfsBmBgsBsin30°

Substitute 200 N for P, 1 m for sA, 5 kg for mA, 9.81m/s2 for g, 25.487 N for Ff, 0.6 m for sB, and 15 kg for mB.

U12=(200×1)+(5×9.81×1)(25.487×0.60)(15×9.81×0.60×sin30°)=200+49.0515.292244.145=189.6128J

Write the equation of work and energy for the system using the equation.

T1+U12=T2

Substitute 0 for T1, 189.6128 J for U12, and 8.272vA2 for T2

0+189.6128=8.272vA2vA=4.7877m/s

Therefore, the velocity of the cylinder A when it strikes the ground is 4.79m/s_.

(b)

Expert Solution
Check Mark
To determine

Find the total distance the block B moves before coming to rest.

Answer to Problem 17.14P

The total distance travelled by the block B before coming to rest is 1.936m_.

Explanation of Solution

Given information:

The mass of the cylinder A is mA=5kg.

The mass of the block B is mB=15kg.

The radius of the outer pulley is rA=250mm.

The radius of the inner pulley is rB=150mm.

The centroidal radius of gyration is kC=160mm.

The coefficient of friction between the surface and the block B is μk=0.20.

The constant force applied at cylinder A is P=200N.

Calculation:

Refer part (a) for vA calculation.

Substitute 4.7877m/s for vA and 250 mm for rA in Equation (1).

ωC=4.7877250mm×1m1,000mm=19.1508rad/s

Substitute 150 mm for rB, 4.7877m/s for vA, and 250 mm for rA in Equation (2).

vB=150×4.7877250=2.87262m/s

Find the total kinetic energy (T3) when the cylinder A strikes the ground.

T3=12mBvB2+12I¯CωC2

Substitute 5 kg for mA, 2.87262m/s for vB, 15 kg for mB, 0.384kgm2 for I¯C, and 19.1508rad/s for ωC.

T3=12×15×2.872622+12×0.384×19.15082=61.89+70.4166=132.3066J

At the final position, the system comes at rest.

The kinetic energy at rest is zero.

T4=0

Apply the principle of work and energy for the block B as follows;

U34=FfsBmBgsBsin30°

Here, the additional distance travelled by the block is sB.

Substitute 9.81m/s2 for g, 25.487 N for Ff, and 15 kg for mB.

U34=25.487×sB15×9.81×sBsin30°=99.062sB

Write the equation of work and energy for the system using the equation.

T3+U34=T4

Substitute 132.3066 J for T3, 99.062sB for U34, and 0 for T4

132.306699.062sB=0sB=1.3356m

Find the total distance (dB) travelled by the block B before coming to rest using the relation.

dB=sB+sB

Substitute 0.6 m for sB and 1.3356 m for sB.

dB=0.6+1.3356=1.9356m

Therefore, the total distance travelled by the block B before coming to rest is 1.936m_.

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Chapter 17 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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