A block rests on a fixed rough inclined plane and a horizontal force F is applied to it, as shown in the figure. Which of the following statements are correct? (1) Normal reaction on the block is Fsin θ +mgcos ⁡ θ (2) The frictional force is zero when Fcos θ =mgsin ⁡ θ (3) The value of limiting friction is μ (.mgsin θ +Fcos ⁡ θ .) (4) The value of limiting friction is μ (.mgsin θ -Fcos ⁡ θ .)

Question

A block rests on a fixed rough inclined plane and a horizontal force F is applied to it, as shown in the figure. Which of the following statements are correct? (1) Normal reaction on the block is Fsin θ +mgcos ⁡ θ (2) The frictional force is zero when Fcos θ =mgsin ⁡ θ (3) The value of limiting friction is μ (.mgsin θ +Fcos ⁡ θ .) (4) The value of limiting friction is μ (.mgsin θ -Fcos ⁡ θ .) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-j0tmf8rlck85.png" /> (A) 1, 2 (B) 3, 4 (C) 2, 4 (D) 2, 3

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-j0tmf8rlmzfx.png" /> As the object is at rest, therefore from FBD, Normal Reaction N=mgcos θ +Fsin ⁡ θ .....(i) Limiting Friction fL=μ N=μ (m g cos θ + F sin ⁡ θ ) ......(ii) Also, if mgsin θ =Fcos ⁡ θ then the object has no tendency of motion; hence friction is zero in this case.

1, 2

3, 4

2, 4

2, 3

As the object is at rest, therefore from FBD, Normal Reaction N=mgcosθ+Fsin⁡θ .....(i) Limiting Friction fL​=μN=μ(mgcosθ+Fsin⁡θ) ......(ii) Also, if mgsinθ=Fcos⁡θ then the object has no tendency of motion; hence friction is zero in this case.

As the object is at rest, therefore from FBD,
Normal Reaction $N = m g cos θ + F s in θ$ .....(i)
Limiting Friction $f_{L} = μ N = μ (m g cos θ + F s in θ)$ ......(ii)
Also, if $m g s in θ = F cos θ$
then the object has no tendency of motion; hence friction is zero in this case.