Asked by Brette

A block of mass 3m is placed on a frictionless
horizontal surface, and a second block of mass m is placed on top of the first
block. The surfaces of the blocks are rough. A constant force of magnitude F
is applied to the first block as shown in the figure.
(a) Construct free-body diagrams for each block.
(b) Identify the horizontal force that causes the block of mass m to accelerate.
(c) Assume that the upper block does not slip on the lower block, and find
the acceleration of each block in terms of m and F.
Answered by bobpursley
the force that causes the upper block to move is friction.

total acceleration:
F=(m+3m)a solve for a. That is the acceleration of both blocks.
Answered by Naseeba
33m.s
Answered by EVE
b) Kinetic Friction
c) Block 3m: F/3m=a
Block m: F/m=a
Answered by Jem
The man pushes with a force of 200 N. The child and sled combo has a mass of 30 kg and the coefficient of static friction is 0.15. Find the friction force
Answered by Jem
Given the figure shown below, block A has a mass of 10kg and block B has a mass of 25kg. Assume the coefficient of friction between the blocks is four times the coefficient of friction between the block and the plane. Determine the normal force between block A and B.
Answered by Bot
The friction force will be equal to the force exerted by the man only when the sled and child start moving, which means we are dealing with static friction. The maximum static friction force is given by:

f_s = μ_s * N

where μ_s is the coefficient of static friction and N is the normal force between the sled and the ground. The normal force is equal to the weight of the sled and child:

N = m*g = 30 kg * 9.81 m/s^2 = 294.3 N

where g is the acceleration due to gravity.

Substituting the given coefficient of friction, we get:

f_s = 0.15 * 294.3 N ≈ 44.15 N

Therefore, the friction force between the sled and the ground is approximately 44.15 N.