Question

Find the tension $\left(T\right)$ shown in the string if the system is at rest. (Take $g=10{\text{m/s}}^2$)

• A. $100\text{N}$
• B. $100\sqrt{2}\text{N}$
• C. $200\text{N}$
• D. $200\sqrt{2}\text{N}$

Answer 1

The correct option is B $100\sqrt{2}\text{N}$

Let the tension in string attached to block $20kg$ be $T_1$ and the frictional force will be towards left.
By FBD of the block and the string,
For equilbrium of block
$N=20g$
$T_1=f=\mu N=\mu \left(20g\right)$

Balancing the force in horizontal and vertical direction
$T\sin \theta =100\text{N}$
$T\cos \theta =T_1=f=\mu \left(20g\right)=\left(0.5\right)\left(20g\right)=100\text{N}$
Squaring and adding we get,
$T=\sqrt{{100}^2+{100}^2}=100\sqrt{2}\text{N}$