Question

A block of mass $10\mathrm{Kg}$ is kept on a rough inclined plane as shown in figure. The coefficient of friction between the block and the surface is $0.6$. Two forces of magnitudes $3\mathrm{N}$ & $\mathrm{P}$ Newton are acting on the block as shown figure. If friction on the block is acting upwards then minimum value of $\mathrm{P}$ for which the block remains at rest is:

• A. $64\mathrm{N}$
• B. $32\mathrm{N}$
• C. $12\mathrm{N}$
• D. $3\mathrm{N}$

Answer 1

The correct option is B

$32\mathrm{N}$

Step 1: Given data and diagram

Mass of a block, $m=10\mathrm{Kg}$

Coefficient of friction, $\mu =0.6$

Now according to the given diagram, if we resolve the normal force, then:

The total upward force will be, $\mathrm{P}+\mathrm{\mu mgcos}{\left(45\right)}^0$

The total downward force will be, $3+\mathrm{mgsin}{\left(45\right)}^0$

Step 2: Calculating the value of $\text{'}\mathrm{P}\text{'}$

In order to put the block at rest, the upward force should be equal to downward force.

$$\begin{gathered} \mathrm{upward}\mathrm{force}=\mathrm{downward}\mathrm{force} \\ \Rightarrow \mathrm{P}+\mathrm{\mu mgcos}{\left(45\right)}^0=3+\mathrm{mgsin}{\left(45\right)}^0 \\ \Rightarrow \mathrm{P}+(0.6\times 10\times 10\times \frac{1}{\sqrt{2}})=3+(10\times 10\times \frac{1}{\sqrt{2}}) \\ \Rightarrow \mathrm{P}+\frac{60}{\sqrt{2}}=3+\frac{100}{\sqrt{2}} \\ \Rightarrow \mathrm{P}=3+\frac{100}{\sqrt{2}}-\frac{60}{\sqrt{2}} \\ \Rightarrow \mathrm{P}=3+\frac{40}{\sqrt{2}} \\ \Rightarrow \mathrm{P}=32\mathrm{N} \end{gathered}$$

Hence, option(B) is correct answer.