two crates, of mass 75 kg and 113 kg, are in contact and at rest on a horizontal surface. A 727 N force is exerted on the 75 kg crate.
If the coefficient of kinetic friction is 0.15, calculate the following:
a)the acceleration of the system
b) the force that each crate exerts on the other
Does the force push the 75 kg crate in the direction of the 113 kg crate, thereby moving both of them? That should have been made clear.
If so, the total friction force is
Ff1 + Ff2 = (75+113)*(0.15)*g = 276 N
The net force accelerating both crates is then
Fnet = 727 - 276 = 451 N
and they both accelerate together at a rate
a = Fnet/(75 + 113) = 2.4 m/s^2
There is a force F' that the two crates exert on each other. You can get it by considering the freebody diagram for either crate. For the 75 kg crate, denoted by "1",
727 N - Ff1 - F' = M1*a = 75*2.4 = 180 N
Ff1 = 0.15*Ma*g = 110 N
F' = 727 -110 -180 = ___
To calculate the acceleration of the system, we need to first determine the net force acting on the system. The net force is the difference between the applied force and the frictional force acting on the crates.
Let's calculate the frictional force first. The frictional force can be represented as the product of the coefficient of kinetic friction (μk) and the normal force (N). Since the crates are at rest on a horizontal surface, the normal force is equal to the weight of the crates.
For the 75 kg crate:
Normal force (N1) = weight (mass1 * gravity)
N1 = 75 kg * 9.8 m/s² = 735 N
For the 113 kg crate:
Normal force (N2) = weight (mass2 * gravity)
N2 = 113 kg * 9.8 m/s² = 1107.4 N
Now, we can calculate the frictional force using the formula:
Frictional force = coefficient of kinetic friction * normal force
For the 75 kg crate:
Frictional force1 = 0.15 * 735 N = 110.25 N
For the 113 kg crate:
Frictional force2 = 0.15 * 1107.4 N = 166.11 N
The net force acting on the system can be calculated as the difference between the applied force and the sum of the frictional forces:
Net force = Applied force - Frictional force1 - Frictional force2
Net force = 727 N - 110.25 N - 166.11 N
Net force = 450.64 N
Since we have the net force, we can now calculate the acceleration of the system using Newton's second law of motion:
Net force = mass_total * acceleration
For the system, the total mass (mass_total) is the sum of the masses of the two crates:
mass_total = mass1 + mass2
mass_total = 75 kg + 113 kg
mass_total = 188 kg
Now we can calculate the acceleration:
450.64 N = 188 kg * acceleration
acceleration = 450.64 N / 188 kg
acceleration ≈ 2.396 m/s²
a) The acceleration of the system is approximately 2.396 m/s².
To calculate the force that each crate exerts on the other, we need to consider Newton's third law: "For every action, there is an equal and opposite reaction." The force that each crate exerts on the other is equal in magnitude but opposite in direction.
Since the system is accelerating, the forces exerted by the crates on each other are not equal to their weights. To calculate the forces, we need to consider the net force acting on each crate separately.
For the 75 kg crate:
Net force1 = Frictional force1 - Applied force
Net force1 = 110.25 N - 727 N
Net force1 = -616.75 N
For the 113 kg crate:
Net force2 = Frictional force2
Net force2 = 166.11 N
Therefore, the force that the 75 kg crate exerts on the 113 kg crate is -616.75 N (opposite to the direction of motion), and the force that the 113 kg crate exerts on the 75 kg crate is 166.11 N (opposite to the direction of motion).
b) The force that each crate exerts on the other is approximately -616.75 N and 166.11 N, respectively.