College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
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Chapter 4, Problem 2P
To determine

The magnitude of force F3 and its angle.

Expert Solution & Answer
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Answer to Problem 2P

The magnitude of force F3 is 78N and the angle is 9.8°.

Explanation of Solution

The figure 1 shows the free body diagram for the particle and the x and y component of three forces

College Physics, Volume 1, Chapter 4, Problem 2P

Write the expression for x component of the 50N force.

  F50N,x=F1cosθ1        (I)

Here, F50N,x is the x component of the 50N force, θ1 is the angle, and F1 is the force of the particle.

Write the expression for y component of the 50N force.

  F50y=F1sinθ1        (II)

Here, F50N,y is the y component of the 50N force.

Write the expression for x component of the 60N force.

  F60N,x=F2cosθ2        (III)

Here, F60N,x is the x component of the 60N force and F2 is the force of the particle.

Write the expression for y component of the 60N force.

  F60N,y=F2sinθ2        (IV)

Here, F60N,y is the y component of the 60N force.

According to static equilibrium, the sum of the x component of the force acting on the particle is,

  Fx=0F50N,x+F60N,x+F3x=0        (V)

Here, F3x is the unknown force acting on the particle along x direction.

According to static equilibrium, the sum of the y component of the force acting on the particle is,

  Fy=0F50N,y+F60N,y+F3y=0        (VI)

Here, F3y is the unknown force acting on the particle along y direction.

The net force F3 acting on the particle is,

  F3=F3x2+F3y2        (VII)

The angle acting on the force F3 is,

  θ3=tan1(F3yF3x)        (VIII)

Here, θ3 is the angle due to the force F3.

Conclusion:

Substitute 50N for F1 and 60° for θ1 in equation (I).

  F50N,x=(50N)cos(60°)=25N

Substitute 50N for F1 and 60° for θ1 in equation (II).

  F50N,y=(50N)sin(60°)=43.3N

Substitute 60N for F2 and 30° for θ2 in equation (III).

  F60N,x=(60N)cos(30°)=52N

Substitute 60N for F2 and 30° for θ2 in equation (IV).

  F60N,y=(60N)sin(30°)=30N

Substitute 25N for F50N,x and 52N for F60N,x in equation (V) to find F3x.

  25N+52N+F3x=077N+F3x=0F3x=77N

Substitute 43.3N for F50N,y and 30N for F60N,y in equation (VI) to find F3y.

  43.3N30N+F3y=013.3N+F3y=0F3y=13.3N

Substitute 77N for F3x and 13.3N for F3y in equation (VII) to find F3.

  F3=(77N)2+(13.3N)2=78N

Substitute 77N for F3x and 13.3N for F3y in equation (VIII) to find θ3.

  θ3=tan1(13.3N77N)=9.8°

Therefore, the magnitude of force F3 is 78N and the angle is 9.8°.

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Chapter 4 Solutions

College Physics, Volume 1

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