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Question: A block \(A\) of mass \(100Kg\) rests on another block \(B\) of \(200Kg\) and is tied to a wall as s...

A block AA of mass 100Kg100Kg rests on another block BB of 200Kg200Kg and is tied to a wall as shown in the figure. The coefficient of friction between AA and BB is 0.20.2 and that between BB and ground is 0.30.3. The minimum force FF required to move the block BB is

A.)900N900N
B.)200N200N
C.)1100N1100N
D.)700N700N

Explanation

Solution

Hint- For getting the unknown forces we should draw a free body diagram to resolve the forces and to get the clear idea of all the forces which are acting on each body separately. So here we will draw a diagram to resolve the forces.

Step By Step Answer:
Here in the diagram we see that there are so many forces like,
Friction force between block AAand block BB = f1{f_1}
Friction force between block BB and the ground = f2{f_2}
Tension in the rope= TT

Minimum force required to move the block BB=FF

Given,

Mass of the block AA=mA={m_A} = 100Kg100Kg
Mass of block BB=mB={m_B} = 200Kg200Kg
Coefficient of friction between AAand BB=μ1={\mu _1} = 0.20.2
Coefficient of friction between BBand ground=μ2={\mu _2} = 0.30.3

Frictional force acting between blocks A and B=f1=μ1mAg=0.2×100×10 = {f_1} = {\mu _1}{m_A}g = 0.2 \times 100 \times 10

f1=200N \Rightarrow {f_1} = 200N----equation (1)

Frictional force acting between block B and ground==f2=μ2mBg=0.3×(100+200)×10 = {f_2} = {\mu _2}{m_B}g = 0.3 \times (100 + 200) \times 10

f2=900N \Rightarrow {f_2} = 900N------equation (2)

Now from the diagram we see that the minimum force that will be required to move the block BB, will be such that it can overcome the friction forces( friction force between both blocks and the friction force between the ground and the block AA). So, for minimum value it must be equal to the sum of both friction forces.

Hence the force required to move the block AA
F=f1+f2F = {f_1} + {f_2} ------ equation (3)

Now putting the values from the equation (1) and equation (2) in the equation (3), we get
F=200+900=1100NF = 200 + 900 = 1100N

Hence the required force will be 1100N1100N.

So, option (C) is the correct answer.

Note- Here the application of the friction force is there. Here we are experiencing two types of friction; one is static friction and another is kinetic friction. The friction which is keeping the block at rest is static friction, and after overcoming the static friction (when the block will start to move) the kinetic friction will come into action.