Question

What is the magnitude of minimum force $F$ necessary to push the block up against a wall, as shown in the figure. The coefficient of friction between the block and the wall is $0.5$, the weight of the block is $12\text{N}$.

• A. $20\text{N}$
• B. $28\text{N}$
• C. $24\text{N}$
• D. $30\text{N}$

Answer 1

The correct option is C $24\text{N}$

By the free body diagram of the block,

condition for equilibrium is,
$F\sin {53}^{\text{o}}=\mu N+mg$,
where weight of the block is given as $12\text{N}$, and $\mu =0.5$

Balancing the forces in horizontal direction
$N=F\cos {53}^{\text{o}}$

By substituting the value of $N$ in the above equation we get,
$F=\frac{mg}{\sin {53}^{\text{o}}-\mu \cos {53}^{\text{o}}}=\frac{12}{\frac{4}{5}-0.5\times \frac{3}{5}}$

$\Rightarrow F=24\text{N}$

Hence option C is the correct answer.