SATHEE: Laws Of Motion Question 401

Laws Of Motion Question 401

Question: Mass of block A is 10 kg. Coefficient of friction between the block A and surface is 0.3. What is the maximum values of M for which system remains in equilibrium $(g = 10 m / s^{2})$

Options:

A) 2 kg

B) 6 kg

C) 3 kg

D) 5 kg

Show Answer

Answer:

Correct Answer: C

Solution:

Maximum friction force between block A and surface is:

$f_{_{max}} = μ m g = (0.3) (10) (10) = 30 N$

Therefore, maximum tension in the string connected with block A should be 30 N.

Now,

$T^{'} \cos 45^{\circ} = 30$ …(1)

$T^{'} s i n 45^{\circ} = M g$ ….(2)

From equations (1) and (2), M = 3 kg

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Laws Of Motion Question 401

Question: Mass of block A is 10 kg. Coefficient of friction between the block A and surface is 0.3. What is the maximum values of M for which system remains in equilibrium(g=10m/s2)

Options:

Answer:

Solution:

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Question: Mass of block A is 10 kg. Coefficient of friction between the block A and surface is 0.3. What is the maximum values of M for which system remains in equilibrium $(g = 10 m / s^{2})$