Question

A $6.5m$ long ladder rests against a vertical wall reaching a height of $6m$. A $60kg$ man stands half way the ladder. The torque of the force exerted by the man on the ladder about the upper end of the ladder (Assuming the weight of the ladder to be negligible as compared to the man and assuming the wall to smooth)

• A. $750N-m$
• B. $850N-m$
• C. $950N-m$
• D. none of these

Answer 1

Answer: (A)

$750N-m$

REF.Image.

Solution

We know that torques magnitude

is equal to the produced of

force and separation of

its line of action and the

point of consideration. so

$\left|{\overrightarrow{T}}_{mg/A}\right|=\frac{Mgb}{2}$

By Pythagoras theorem

$(6.5{)}^2=(6{)}^2+(6{)}^2$

$\Rightarrow b=\sqrt{(6.5{)}^2-(6{)}^2}=2.5$

So, $\left|{\overrightarrow{T}}_{mg/A}\right|=\frac{\left(60\right)\left(10\right)2.5}{2}=-750Nm$

option - A is correct.