Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. Fig. P2.127

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Answer 1

Textbook Question

Chapter 2, Problem 2.127RP

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Chapter 2, Problem 2.127RP, Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q =

Fig. P2.127

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

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Answer 2

Textbook Question

Chapter 2, Problem 2.127RP

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Chapter 2, Problem 2.127RP, Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q =

Fig. P2.127

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 2, Problem 2.127RP

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Chapter 2, Problem 2.127RP, Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q =

Fig. P2.127

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 2, Problem 2.127RP

Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.

Chapter 2, Problem 2.127RP, Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q =

Fig. P2.127

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Answer 8

Expert Solution & Answer

To determine

The direction and magnitude of the resultant of two forces.

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The direction and magnitude of the resultant of two forces.

Answer 12

Answer to Problem 2.127RP

The magnitude of the resultant of two forces is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Answer 13

Explanation of Solution

The sketch of the triangle formed by the two forces $P$ , $Q$ , and resultant $R$ is shown in figure $1$ .

VECTOR MECHANICS FOR ENGINEERS: STATICS, Chapter 2, Problem 2.127RP

Write the expression for the total angle of a triangle

$α+ϕ+γ=180°$ (I)

Here, $α$ , $ϕ$ , and $γ$ are the three angles of a triangle.

Write the expression for the angle $ϕ$ formed at the base

$ϕ=180°−α−80°$ (II)

The sum of the three angles of a triangle is equal to $180°$ .

Write the expression for cosine law for finding $R$ in figure

$R2=A2+B2−2(ABcosγ)$ (III)

Here, $R$ is the hypotenuse which is representing resultant of two forces $P$ and $Q$

Write the expression for law of Sines

$Asinα=Bsinϕ=Rsinγ$ (IV)

Here, $A$ , $B$ , and $C$ are the side lengths of triangle.

Conclusion:

Use equation (I) to obtain the angle $γ$ as shown in figure

$γ=180°−(20°+10°)=150°$

Substitute $48 N$ for $A$ , $60 N$ for $B$ , and $150°$ for $γ$ to obtain resultant force $R$

$R2=(48 N)2+(60 N)2−2(48 N×60 N×cos150°)=10892.3R=104.366 N$

Use the law of sines to find the angle $α$ .

Substitute $48 N$ for $A$ , $104.366 N$ for $R$ , and $150°$ for $γ$ in equation (III) to find $α$

$48 Nsinα=104.366 Nsin150° Nsinα=48 N×sin150° 104.366 N=0.22996α=13.2947°$

Substitute $13.2947°$ for $α$ in equation (II)

$ϕ=180°−13.2947°−80°=86.705°$

Therefore the magnitude of resultant force is $R=104.4 N_$ , and the direction is $ϕ=86.7°_$ .

Answer 14

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Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces. Fig. P2.127

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