Two blocks of masses m = 5 kg and M = 10 kg are connected by a string passing over a pulley B as shown. Another string connects the centre of pulley B to the floor and passes over another pulley A as shown. An upward force F is applied at the centre of pulley A. Both the pulleys are massless.
Find the acceleration of blocks m and M, if F is
(a) 100 N
(b) 300 N
(c) 500 N (Take g = 10 m/s²)

Question

Accepted Answer

Let [latex]T_{0}[/latex] = tension in the string passing over A
T = tension in the string passing over B

[latex]\begin{aligned} 2 T_0 &=F \quad ext { and } \quad 2 T=T_0 \ T &=F / 4\end{aligned}[/latex]

(a) T=F/4=25 N

weights of blocks are [latex]\begin{aligned} m g &=50 \mathrm{~N} \ M g &=100 \mathrm{~N} \end{aligned}[/latex]

AsT mg < and Mg both, the blocks will remain stationary on the floor.

(b) T=F/4=75 N

As T Mg < and T >, mg M will remain stationary on the floor, whereas m will move.
Acceleration of m,

[latex]\begin{aligned} a &=\frac{T-m g}{m}=\frac{75-50}{5} \ &=5 \mathrm{~m} / \mathrm{s}^2 \end{aligned}[/latex]

(c) T=F/4-125 N

As T mg > and Mg both the blocks will accelerate upwards.
Acceleration of m,

[latex]a_1=\frac{T-m g}{m}=\frac{125-50}{5}=15 \mathrm{~m} / \mathrm{s}^2[/latex]

Acceleration of M,

[latex]a_2=\frac{T-M g}{M}=\frac{125-100}{10}=2.5 \mathrm{~m} / \mathrm{s}^2[/latex]

Question 8.SE.19: Two blocks of masses m = 5 kg and M = 10 kg are connected by...

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