Question

the minimum value of f such that the block remains at rest

Answer 1

Answer:

c

Explanation:

First draw fbd of the block.

N=Fcos theta

Fsine theta+friction=mg

Fsine theta+mew*N=mg

as N=fcos theta

F=mg/sine theta+mew cos theta

And option d is for the maximum value of force...

THANK YOU

Answer 2

Answer:

For the block to remain at rest, the minimum value of force $F$ is $\frac{mg}{(sin\theta +\mu cos\theta)}$. Option (c) is correct.

Explanation:

• If the net horizontal and vertical forces are equal to zero, the block will remain at the rest.
• The external force and the frictional force will act in the opposite direction to the weight of the body.

Step 1:

In the horizontal direction:

$Fcos\theta=R$

where $R$ is the reaction provided by the surface

Step 2:

In the vertical direction:

$Fsin\theta +\mu R=mg$

Substituting the value of reaction $R$ we get:

$Fsin\theta +\mu (Fcos\theta)=mg$

$F(sin\theta +\mu cos\theta)=mg$

$F=\frac{mg}{(sin\theta +\mu cos\theta)}$

So, for the block to remain at rest, the minimum value of force $F$ is $\frac{mg}{(sin\theta +\mu cos\theta)}$.

Option (c) is correct.