Question

The rough surface of a horizontal table
offers a definite maximum opposing force
to initiate the motion of a block along the
table, which is proportional to the resultant normal force given by the table. ForcesF1, and F2, act at the same angle 0 with the horizontal and both are just initiating the sliding motion of the block along the table. Force F1, is a pulling force while the force F2, is a pushing force. F1 > F2 , because
(A) Component of F, adds up to weight to
increase the normal reaction.
(B) Component of F, adds up to weight to
increase the normal reaction.
(C) Component of F, adds up to the
opposing force.
(D) Component of F, adds up to the
opposing force.​

Answer 1

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Component of F2, adds up to weight to  increase the normal reaction.

Option (A) is correct. ✔

Explanation:

We can assume that the frictional force is acting along the surface of the table while the weight is acting in a downward direction.  

Now mu (μ) = mg

Twh frictional force is fs (max) = μ mg = μ H

fs (max)  ∝ μ

As the component of force is making an angle theta along the surface therefore the coefficient of friction will be  

μ = mg F2sin θ ✔

Answer 2

Answer:

Component of F2, adds up to weight to increase the normal reaction.

Option (A) is correct. ✔

Explanation:

We can assume that the frictional force is acting along the surface of the table while the weight is acting in a downward direction.

Now mu (μ) = mg

Twh frictional force is fs (max) = μ mg = μ H

fs (max) ∝ μ

As the component of force is making an angle theta along the surface therefore the coefficient of friction will be

μ = mg F2sin θ ✔