Question

Say you have a scenario set up where someone pushed 2 blocks toward you and then let go so you no longer have an applied force from their direction. One being Block A at 2kg and the other being Block B at 4 kg. Block A is on the side closer to you and Block B is touching the other side of Block A, which is what the other person pushed on. The blocks are made of the same material and both have kinetic coefficients of μk=0.12. Now while the blocks are sliding toward you, you apply a force with your hand on block A, that applies an acceleration of 2m/s^2 to the blocks. Since the blocks are sliding towards you, that means that kinetic friction is working with your hand force to slow the blocks down. Which means that the friction is part of the 2m/s^2 acceleration, along with your hand. Now the problem I am having is calculating the contact force between the two blocks during this scenario. I know that when the friction is opposing your applied force, you would subtract the kinetic friction of block B from its (mass)*(acceleration) and that would give you the contact force between the two blocks, but since the friction is working with you in this scenario does that mean you do not have to subtract it? I might have said some things wrong in this explanation because I am new to this. If you need more clarification let me know.

Answer 1

Refer to the diagram:

나 도 TY 2

given:

$\mu_{k}=0.12$

mA = mass of A = 2kg

mB = mass of B = 4kg

a = acceleration = $2m/s^{2}$

N = contact force between the blocks.

Left block is A and the right block is B.

The blocks are moving towards left causing friction towards right, the force exerted by the person towards right.

Consider the block B:

friction force on B =

f2 = k тву

Applying Newton's 2nd law (The direction of force in the direction of acceleration is taken positive, and the force opposite to the direction of motion is taken negative, then add the forces to give mass*acceleration) :

$F_{\text{net on B}} = m_{B}a$

$\Rightarrow N+f_{2} = m_{B}a$

$\Rightarrow N+\mu_{k}m_{B}g = m_{B}a$

$\Rightarrow N = m_{B}a-\mu_{k}m_{B}g$

$\Rightarrow N = m_{B}(a-\mu_{k}g)=4(2-0.12*9.18)=3.59N$

Therefore, the contact force between the blocks is 3.59N [answer]