A 4kg block and a 2kg block can move on horizontal frictionless surface. The blocks are accelerated by a +12-N force that pushes
the large block against the smaller one.'
a) Determine the force that the 2kg block exerts on the 4kg block
b) Determine the force that the 4kg block exerts on the 2kg block.
a) Determine the force that the 2kg block exerts on the 4kg block
b) Determine the force that the 4kg block exerts on the 2kg block.
1 answer:
Answer:
a) -4 N
b) +4 N
Explanation:
Draw a free body diagram for each block.
For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.
For the small block, there is 1 force, F pushing to the right.
There are also weight and normal forces in the vertical direction, but we can ignore those.
Sum of forces on the large block in the x direction:
∑F = ma
12 − F = 4a
Sum of forces on the small block in the x direction:
∑F = ma
F = 2a
2F = 4a
Substitute:
12 − F = 2F
12 = 3F
F = 4
The small block pushes on the large block 4 N to the left (-4 N).
The large block pushes on the small block 4 N to the right (+4 N).
You might be interested in
Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
