Question

In the arrangement shown, $W_1=2000N,W=100N,\mu =0.25$ for all surfaces in contact. The block $W_1$ just slides under the block $W_2$then what will be the tension in the string.

• A. A pull of 50 N is to be applied on $W_1$
• B. A pull of 90 N is to be applied on $W_1$
• C. Tension in the string AB is 10 N
• D. Tension in the string AB is $20\sqrt{2}N$

Answer 1

The correct option is D Tension in the string AB is $20\sqrt{2}N$

FBD will as

Let P will be max pull for no sliding

Then,

$N+\frac{T}{\sqrt{2}}={\omega }_2$

$f_1=\frac{T}{\sqrt{2}}$

$f_1=\mu N=N/4$

$P=f_1+f_2$

$f_2=\mu (N+{\omega }_1)=\frac{1}{4}(N+{\omega }_1)$

From these equation, we get

$\Rightarrow \frac{N}{4}+N={\omega }_2\Rightarrow N=\frac{4}{5}\left(100\right)=80N$

$T=\frac{N}{2\sqrt{2}}=\frac{80}{2\sqrt{2}}=20\sqrt{2}N$

$P=20N+\frac{1}{4}(80+2000)$

$P=20+520=540$N

Option D is correct.