PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
PHYSICS F/SCI.+ENGRS.,STAND.-W/ACCESS
6th Edition
ISBN: 9781429206099
Author: Tipler
Publisher: MAC HIGHER
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Chapter 1, Problem 77P

(a)

To determine

To find: The two forces in unit vectors.

(a)

Expert Solution
Check Mark

Answer to Problem 77P

The force applied by person P is (35.36lb)i^+(35.36lb)j^ and the force applied by Person Jis (53.24lb)i^(37.28lb)j^ .

Explanation of Solution

Given:

The force applied by person P is 50lb in northeast direction.

The force applied by person J is 65lb , 35° south of due west.

Formula used:

Write the expression for the horizontal component of force.

  Fx=Fcosθ

Write the expression for the vertical component of force.

  Fy=Fsinθ

Write the expression of force in terms of a unit vector.

  FR=Fxi^+Fyj^ …… (1)

Substitute Fcosθ for Fx and Fsinθ for Fy in equation (1).

  FR=Fcosθi^+Fsinθj^ …… (2)

Here, F is the magnitude of the force, FR is the force in vector form and θ is the angle.

Calculation:

For person P:

The direction of force is between the positive x-axis and the positive y-axis.

Substitute 50lb for F and 45° for θ in equation (2).

  F1=(50lb)cos45°i^+(50lb)sin45°j^=(35.36lb)i^+(35.36lb)j^

For Person J:

The direction of force is between the negative x-axis and the positive y-axis.

Substitute 65lb for F and 35° for θ in equation (2).

  F2=(65lb)cos35°i^(65lb)sin35°j^=(53.24lb)i^(37.28lb)j^

Conclusion:

Thus, the force applied by person P is (35.36lb)i^+(35.36lb)j^ and the force applied by Person Jis (53.24lb)i^(37.28lb)j^ .

(b)

To determine

To write: The force applied by person C.

(b)

Expert Solution
Check Mark

Answer to Problem 77P

The force applied by person C is (17.88lb)i^+(1.92lb)j^ ; magnitude of force 17.98lb at an angle of 6.13° north of due east.

Explanation of Solution

Given:

The resultant of three forces equal to zero.

Formula used:

The force applied by person C should be equal and opposite to the resultant sum of forces applied by person P and person J.

Write the expression for the force applied by person C.

  F3=(F1+F2) …… (3)

Here, F3 is the force applied by person C.

Calculation:

Substitute (35.36lb)i^+(35.36lb)j^ for F1 and (53.24lb)i^(37.28lb)j^ for F2 in equation (3).

  F3=[(35.36lb)i^+(35.36lb)j^(53.24lb)i^(37.28lb)j^]=(17.88lb)i^+(1.92lb)j^

Calculate the magnitude of force F3 .

  |F3|= ( 17.88lb )2+ ( 1.92lb )2=17.98lb

Calculate the angle of force.

  θ=tan1( 1.92lb 17.88lb)=6.13°

Conclusion:

Thus, the force applied by person Cis (17.88lb)i^+(1.92lb)j^ ; the magnitude of force 17.98lb at an angle of 6.13° north of due east.

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