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# using free body diagram show that it is easy to pull an object than to push it

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## Using a free body diagram, show that it is easy to pull an object than to push it.

Using a free body diagram, show that it is easy to pull an object than to push it.

Using a free body diagram, show that it is easy to pull an object than to push it.

### SOLUTION

When a body is pushed at an arbitrary angle θ [0 to

π π2

], the applied force F can be resolved into two components as F sin 0 parallel to the surface and F cos 0 perpendicular to the surface as shown in the figure. The total downward force acting on the body is mg + F cos θ. It implies that the normal force acting on the body increases. Since there is no acceleration along the vertical direction the normal force N is equal to

Npush = mg + F cos θ …………(1)

As a result the maximal static friction also increases and is equal to

mg

fSmax=μrNpush=μs(mg+Fcosθ)

.................(2)

Equation (2) shows that a greater force needs to be applied to push the object into motion.

An object is pushed at an angle θ

When an object is pulled at an angle θ, the applied force is resolved into two components as shown in the figure. The total downward force acting on the object is –

Npull = mg – F cos θ ………….(3)

An object is pulled at an angle θ

Equation (3) shows that the normal force is less than – Npush. From equations (1) and (3), it is easier to pull an object than to push to make it move.

Concept: Application of Newton's Law

Is there an error in this question or solution?

Chapter 3: Laws of motion - Evaluation [Page 162]

Q II. 5. Q II. 4. Q II. 6. a.

### APPEARS IN

Tamil Nadu Board Samacheer Kalvi Class 11th Physics Volume 1 and 2 Answers Guide

Chapter 3 Laws of motion

Evaluation | Q II. 5. | Page 162

स्रोत : www.shaalaa.com

## Using free body diagram show that it is easy to pull an object than to push it

When a body is pushed at an arbitrary angle theta that varies from (p " to " (pi )/(2)) . The applied force F can be resolved into two components as F sin theta parallel to the surface and F cos theta prependicular to the surface . The total downward force acting on the body mg + F cos theta . It is implied that the normal force acting on the body increase . since there is no acceleration along the vertical direction the normal force N is given by As a result the maximal static friction also increases and that is equal to f(s) ^(max) = mu(s) N(push) -mu(s ) (mg + F cos theta) Equation shows that a greater force needs to be applied to push the object into motion When an object is pulled at an angle theta the apllied force is resolved into two components . The total downward force acting on the object is Equation (3) shows that the normal force is less that N(push) From equation (1) and (3) we know that it is easier to pull an object that to push to make it move.

Using free body diagram show that it is easy to pull an object than to push it

PREMIERS PUBLISHERS-LAW OF MOTION -EVALUATION (TEXTBOOK QUESTIONS & ANSWERS) (II SHORT ANSWER QUESTIONS)

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Updated On: 27-06-2022

Text Solution Solution

When a body is pushed at an arbitrary angle

θ that varies from ( p to π 2 )

. The applied force F can be resolved into two components as F sin

θ

parallel to the surface and F cos

θ

prependicular to the surface . The total downward force acting on the body mg + F cos

θ

. It is implied that the normal force acting on the body increase . since there is no acceleration along the vertical direction the normal force N is given by

As a result the maximal static friction also increases and that is equal to

f max s = μ s N p u s h − μ s ( m g + F cos θ )

Equation shows that a greater force needs to be applied to push the object into motion

When an object is pulled at an angle

θ

the apllied force is resolved into two components . The total downward force acting on the object is

Equation (3) shows that the normal force is less that

N p u s h From equation ( 1 )

and (3) we know that it is easier to pull an object that to push to make it move.

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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