Question

The block of mass $m\text{kg}$ is being pulled by a horizontal force $F=2\text{mg}$ applied to a string as shown in figure (Take $g=10{\text{ms}}^{-2})$. The pulley is massless and is fixed at the edge of an immovable table with a supporting cable. What is the value of force exerted by the supporting cable (rod) on the pulley (in $\text{Newton}$).

• A. $2\sqrt{2}mg$
• B. $4mg$
• C. $2mg$
• D. $3mg$

Answer 1

The correct option is A $2\sqrt{2}mg$

FBD of the pulley is shown in the figure:


$F_1$ be the horizontal force exerted by string i.e $F_1=2mg$

Since, the string is same and a massless string transmits the same tension every where.

$F_2$ be the force exerted by vertical part of string on pulley, so $F_2=F_1=2mg$

Let $F_3$ be the force exerted by supporting rod on pulley.

Pulley is massless. Thus net force on it is zero.

$\Rightarrow \overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0$

$\Rightarrow$$\left|\begin{array}{c}\overrightarrow{F_3}\end{array}\right|=\left|\begin{array}{c}\overrightarrow{F_1}+\overrightarrow{F_2}\end{array}\right|$

$\Rightarrow$ $\left|\begin{array}{c}\overrightarrow{F_3}\end{array}\right|=\sqrt{{\left(2mg\right)}^2+{\left(2mg\right)}^2}$

$\therefore \left|\begin{array}{c}\overrightarrow{F_3}\end{array}\right|=2\sqrt{2}mg$

Hence, option (c) is the correct answer.