SATHEE: Laws Of Motion Question 404

Laws Of Motion Question 404

Question: Two blocks (m and M) are arranged as shown in Fig. 21. If there is friction between ground and M only and other surfaces are frictionless. The coefficient of friction between ground and M is $μ = 0.75$ . The maximum ratio of m and M (m/M) so that the system remains at rest is

Options:

A) $1 / 4$

B) 3

C) $1 / 3$

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

As sphere is at rest,

$T = m g$

$f = T$

and $f \leq μ N$

Here $N = M g + T = (M + m) g$

$T \leq m (M + m) g$

$m g \leq μ (M + m) g$

$\frac{m}{M} \leq \frac{μ}{(1 - μ)}$

$\frac{m}{M} \leq \frac{0.75}{(1 - 0.75)} \Rightarrow \frac{m}{M} \leq 3$

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Laws Of Motion Question 404

Question: Two blocks (m and M) are arranged as shown in Fig. 21. If there is friction between ground and M only and other surfaces are frictionless. The coefficient of friction between ground and M is μ=0.75 . The maximum ratio of m and M (m/M) so that the system remains at rest is

Options:

Answer:

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Question: Two blocks (m and M) are arranged as shown in Fig. 21. If there is friction between ground and M only and other surfaces are frictionless. The coefficient of friction between ground and M is $μ = 0.75$ . The maximum ratio of m and M (m/M) so that the system remains at rest is