EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100460300
Author: SERWAY
Publisher: YUZU
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Chapter 5, Problem 5.19P

Two forces F 1 and F 2 act on a 5.00-kg object. Taking F1 = 20.0 N and F2 = 15.0 N, find the accelerations of the object for the configurations of forces shown in parts (a) and (b) of Figure P5.19.

Chapter 5, Problem 5.19P, Two forces F1 and F2 act on a 5.00-kg object. Taking F1 = 20.0 N and F2 = 15.0 N, find the

Figure P5.19

(a)

Expert Solution
Check Mark
To determine

The acceleration of the object for the configurations of forces shown in part (a) of given figure.

Answer to Problem 5.19P

The magnitude of the acceleration is 5m/s2 and the angle made by the acceleration with the horizontal is 36.869° .

Explanation of Solution

Given info: Two forces F1 and F2 act on a object which mass is 5.00kg . The value of F1 is 20.0N and the value of F2 is 15.0N and the angle between F1 and F2 is 90° .

The net force act on the object is,

F=Fxi^+Fyj^ (1)

Here,

Fx is the horizontal component of force.

Fy is the vertical component of force.

The horizontal component of force is,

Fx=F1cosθ1+F2cosθ2

In the horizontal direction the F1 makes an angle 0° and F2 makes an angle 90° .

Substitute 0° for θ1 , 90° for θ2 , 20.0N for F1 and 15.0N for F2 in above equation.

Fx=(20.0N)cos0°+(15.0N)cos90°=20.0N+0=20.0N

The vertical component of force is,

Fy=F1sinθ1+F2sinθ2

In the vertical direction the F1 makes an angle 90° and F2 makes an angle 0° .

Substitute 90° for θ1 , 0° for θ2 , 20.0N for F1 and 15.0N for F2 in above equation.

Fy=(20.0N)sin0°+(15.0N)sin90°=0+15.0N=15.0N

Substitute 20.0N for Fx and 15.0N for Fy in equation (1).

F=(20.0i^+15.0j^)N

According to Newton’s second law of motion,

F=ma

Here,

a is the acceleration of the object.

Substitute (20.0i^+15.0j^)N for F and 5.00kg for m in above equation.

(20.0i^+15.0j^)N=(5.00kg)aa=(20.0i^+15.0j^)N(5.00kg)=(4i^+3j^)m/s2

The magnitude of the acceleration is,

a=ax2+ay2

Here,

ax is the acceleration in horizontal direction.

ay is the acceleration in vertical direction.

Substitute 4 for ax and 3 for ay in above equation.

a=42+32m/s2=(16+9)m/s2=25m/s2=5.00m/s2

The angle made by acceleration with the horizontal is,

tanθ=ayax

Substitute 4 for ax and 3 for ay in above equation.

tanθ=34θ=tan1(34)=36.869°

Conclusion:

Therefore, the magnitude of the acceleration is 5.00m/s2 and the angle made by the acceleration with the horizontal is 36.869° .

(b)

Expert Solution
Check Mark
To determine

The acceleration of the object for the configurations of forces shown in part (b) of given figure.

Answer to Problem 5.19P

The magnitude of the acceleration is 6.08m/s2 and the angle made by the acceleration with the horizontal is 25.3° .

Explanation of Solution

Given info: Two forces F1 and F2 act on a object which mass is 5.00kg . The value of F1 is 20.0N and the value of F2 is 15.0N and the angle between F1 and F2 is 60° .

The horizontal component of force is,

Fx=F1cosθ1+F2cosθ2

In the horizontal direction the F1 makes an angle 0° and F2 makes an angle 60° .

Substitute 0° for θ1 , 60° for θ2 , 20.0N for F1 and 15.0N for F2 in above equation.

Fx=(20.0N)cos0°+(15.0N)cos60°=20.0N+(15.0N)(0.5)=27.5N

The vertical component of force is,

Fy=F1sinθ1+F2sinθ2

In the vertical direction the F1 makes an angle 60° and F2 makes an angle 0° .

Substitute 60° for θ1 , 0° for θ2 , 20.0N for F1 and 15.0N for F2 in above equation.

Fy=(20.0N)sin0°+(15.0N)sin60°=0+(15.0N)(0.866)=13.0N

Substitute 27.5N for Fx and 13.0N for Fy in equation (1).

F=(27.5i^+13.0j^)N

According to Newton’s second law of motion,

F=ma

Substitute (27.5i^+13.0j^)N for F and 5.00kg for m in above equation.

(27.5i^+13.0j^)N=(5.00kg)aa=(27.5i^+13.0j^)N(5.00kg)=(5.5i^+2.6j^)m/s2

The magnitude of the acceleration is,

a=ax2+ay2

Substitute 5.5 for ax and 2.6 for ay in above equation.

a=(5.5)2+(2.6)2m/s2=(30.25+6.76)m/s2=37.01m/s2=6.08m/s2

The angle made by acceleration with the horizontal is,

tanθ=ayax

Substitute 5.5 for ax and 2.6 for ay in above equation.

tanθ=5.52.6θ=tan1(5.52.6)=tan1(2.115)=25.3°

Conclusion:

Therefore, the magnitude of the acceleration is 6.08m/s2 and the angle made by the acceleration with the horizontal is 25.3° .

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Chapter 5 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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