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.
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.
b) Kinetic Friction
c) Block 3m: F/3m=a
Block m: F/m=a
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
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.
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.
(a) To construct the free-body diagrams for each block, we need to identify all the forces acting on them.
For the first block of mass 3m:
1. Weight (mg) acting downward - This force can be represented as a downward arrow.
2. Normal force (N1) exerted by the horizontal surface - This force acts perpendicular to the surface and can be represented as an upward arrow.
3. Force of tension (T) exerted by the second block - This force acts horizontally and can be represented as an arrow pointing from the first block towards the second block.
For the second block of mass m:
1. Weight (mg) acting downward - This force can be represented as a downward arrow.
2. Force of tension (T) exerted by the first block - This force acts horizontally and can be represented as an arrow pointing from the second block towards the first block.
3. Normal force (N2) exerted by the first block - This force acts perpendicular to the surface and can be represented as an upward arrow.
(b) The horizontal force that causes the block of mass m to accelerate is the force of tension (T) exerted by the first block. This is because the force of tension is the only horizontal force acting on the second block.
(c) To find the acceleration of each block, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
For the first block of mass 3m:
The net force acting on this block is F - T (the applied force minus the force of tension). Since there is no acceleration in the vertical direction, the normal force N1 cancels out the weight (mg). So,
F - T = (3m) * a1
For the second block of mass m:
The net force acting on this block is T (the force of tension). Since there is no acceleration in the vertical direction, the normal force N2 cancels out the weight (mg). So,
T = m * a2
Since the two blocks are connected and have the same acceleration (a1 = a2 = a), we can solve these equations simultaneously:
F - T = (3m) * a
T = m * a
Adding these two equations together, we get:
F = (4m) * a
Solving for acceleration, we find:
a = F / (4m)
So, the acceleration of each block in terms of m and F is a = F / (4m).