Question

A body of weight 150N lies on a horizontal plane for which cofficient of friction is 0.70. determine
1. normal reaction
2. limiting force of friction
3. horizontal force required to move it
4. angle of friction​

Answer 1

Given that,

weight of body = 150 N

cofficient of friction = 0.70

to find,

1.) normal reaction = weight = 150 N

2.) limiting force of friction

= 150 × 0.70

= 105 N

3.) horizontal force required to move it

= limiting force of friction = 105 N

4.) angle of friction = ©

tan© = 0.7

© = tan^-1 (0.7)

Answer 2

Answer:

1. 150N

2.105N

3.105N

4.Tan^-1(0.7)

Explanation:

Given

weight of body = 150 N

cofficient of friction = 0.70

Find

normal reaction = weight = 150 N

limiting force of friction

horizontal force required to move it

angle of friction

Solution

1. normal reaction = weight = 150 N

• limiting force of friction== 150 × 0.70= 105 N
• horizontal force required to move it= limiting force of friction = 105 N
• angle of friction =θ

       Tanθ=0.7

        θ=$tan^{-1}$(0.7)

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