Question

Two blocks (M₁=5 kg and M₂=3 kg) are stacked on a frictionless surface but there is friction between them. The coefficient of static friction between the blocks is μ_s=0.6 and the coefficient of kinetic friction is μ_k=0.3. If a person applies a horizontal force F=60 N to M₁, the blocks move. You do not know if the blocks slip or accelerate together until you check.

M2 ^^ M1 ---> X

1. The friction between M₁ and M₂ is:

2. What is the magnitude of the friction force between the two blocks?

Answer 1

1. To solve this question we assume that friction between $M_{1}$ and $M_{2}$ is static friction so block $M_{2}$ does not slip over block $M_{1}$ when force F is applied to block $M_{1}$ . Let the magnitude of friction force between blocks is 'f' and the acceleration of the blocks is 'a' along +x axis. Now we will find out the condition from which we can get the minimum value of coefficient of static friction interms of $M_{1}$ , $M_{2}$ , F and g ( acceleration due to gravity) and we will compare that value of coefficient of static friction with the given value in the problem.

To find out this minimum value of static friction we use Newtown laws of motion.

First we consider the forces on block $M_{1}$ . There are two force on $M_{1}$ , one is the external applied force F and the other one is the friction force f between blocks. The direction of F is in the +x axis but the direction of friction force f is in the -x axis. Let the acceleration of the blocks is a. Then applying Newton's second law of motion we obtain

F-f= Mia

Similarly for block $M_{2}$ the only force along x axis is the static friction force f. Now the direction of friction force on $M_{2}$ is along the +x axis. So applying Newton's second law of motion we obtain

f = Mga

Now solving above two equations we have

$a=\frac{F}{M_{1}}-\frac{f}{M_{1}}=\frac{f}{M_{2}}$

Ffff(M2 + M2) M.= M M = MM,

FM2 J= Mi + M2

This is the value of static friction.

Now we know that the maximum value of static friction is

marimum = us M29

If this
marimum
is greater than the obtain value from our calculations then the block $M_{2}$ will not slip over $M_{1}$ . So the condition for which, block $M_{2}$ will not slip over the block $M_{1}$ that is the friction between the blocks will be static friction is given below

FM fmarimum Mi+ M2

MsM29 FM2 Mi + M2

Jus Mi + M29

This is the condition from which we can find out the minimum value of coefficient of static friction. Now using the value given in the problem

F = 601

M1 = 5kg

M2 = 3kg

. This is the value of acceleration due to gravity.

g=9.8m/s2

Jus 60 (5+3) x 9.8

р, > 0.7653

But we know from the problem that the coefficient of static friction between blocks is 0.6 which is less than the minimum value 0.7653 for which the block $M_{2}$ will not slip over block $M_{1}$ . So the block $M_{2}$ will start to slip when F is applied to $M_{1}$ and the friction between the blocks is kinetic friction.

so the friction between $M_{1}$ and $M_{2}$ is kinetic friction.

2. Now the magnitude of the friction force can be easily find out because we know that the friction between blocks is kinetic friction. So the magnitude of the friction force between the blocks is

kinetic = fkM29

μα = 0.3

fkinetic = 0.3 x 3 x 9.8 = 8.82N

This is the magnitude of the friction force between the blocks.