Question 119: Laws of Motion

Question

Three masses $$M=100 \mathrm{~kg}, \mathrm{~m}_{1}=10 \mathrm{~kg}$$ and $$\mathrm{m}_{2}=20 \mathrm{~kg}$$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $$\mathrm{F}$$ is applied on the system so that the mass $$\mathrm{m}_{2}$$ moves upward with an acceleration of $$2 \mathrm{~ms}^{-2}$$. The value of $$\mathrm{F}$$ is :
( Take $$\mathrm{g}=10 \mathrm{~ms}^{-2}$$ )

Accepted Answer

Here we will use the Newton's laws of motion.
Let acceleration of 100 kg block w.r.t. ground is a.
Now, Let's draw FBD of 100 kg block w.r.t. ground,
By NLM,
$F-T-N_1=100a\quad 	ext{... (1)}$
FBD of 20 kg block w.r.t. 100 kg block,
By NLM,
$${N_1} = {m_2}a = 20a \Rightarrow {N_1} = 20a\quad$$ ... (2)
$$T - {m_2}g = {m_2}(2)$$
$$ \Rightarrow T - 20g = 20(2)$$
$$ \Rightarrow T = 40 + 200 = 240$$
$$ \Rightarrow T = 240N\quad$$ .... (3)
Now, FBD of 10 kg block w.r.t. 100 kg block,
By NLM,
$${m_1}a - T = {m_1}(2)$$
$$ \Rightarrow 10a - T = 20$$
$$ \Rightarrow 10a - 240 = 20\quad$$ [from (3)]
$$ \Rightarrow 10a = 260$$
$$ \Rightarrow a = 26\,m{s^{ - 2}}$$
So, from (2), $${N_1} = 20(26) = 520N$$
Now, from (1), $$F - 240 - 520 = 100(26)$$
$$ \Rightarrow F = 2600 + 240 + 520$$
$$ \Rightarrow F = 3360\,N$$

Answer

3360 N

Answer

3380 N

Answer

3120 N

Answer

3240 N

Laws of Motion

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