A 6.00 kg crate is pushed by a 300.0 N force at and angle of 30 degrees with the horizontal.
a.What coefficient of friction value is experienced by the crate if it moves at constant velocity?
b. What coefficient of friction value is present if the crate accelerates at 1.50m/s2
The answer to (a) depends upon whether the 30 degree angle of the pushing force is upwards or downwards, with respect to the floor. I will assume downwards.
The weight of the crate is M g = 6.0 x 9.8 = 58.8 N
The total force component downwards on the floor is then 58.8 + 300 sin 30 = 208.8 N
If there is no acceleration, the net horizontal force is zero
300 cos 30 - 208.8 * f = 0
Solve for the coefficient of friction, f
For (b), get the net force from
Fnet = M a = 450 N
Fnet = 300 cos 30 - 208.8 f = 450
Solve for the (new) value of f.
To solve these problems, we need to use Newton's laws of motion and the concept of friction. Here's how you can calculate the coefficient of friction for both scenarios:
(a) If the crate moves at a constant velocity, the net force acting on it is zero. Therefore, the horizontal component of the applied force must balance the frictional force.
Let's break down the forces involved:
- The weight of the crate is M * g, where M represents the mass (6.00 kg) and g is the acceleration due to gravity (9.8 m/s^2). So, the weight is 6.00 kg * 9.8 m/s^2 = 58.8 N.
- The vertical component of the applied force is given by 300 N * sin(30 degrees) = 150 N.
- The net downward force on the crate is the sum of its weight and the vertical component of the force: 58.8 N + 150 N = 208.8 N.
Since the crate moves at a constant velocity, the net horizontal force must be zero. So, the horizontal component of the applied force must balance the frictional force. The horizontal component is given by 300 N * cos(30 degrees) = 300 N * (√3/2) = 259.8 N.
Now we can set up the equation:
259.8 N - 208.8 N * f = 0,
where f represents the coefficient of friction. Rearranging the equation, we get:
f = 259.8 N / 208.8 N ≈ 1.244.
Therefore, the coefficient of friction experienced by the crate when it moves at a constant velocity is approximately 1.244.
(b) If the crate accelerates at 1.50 m/s^2, there is a net horizontal force acting on it. Again, we need to calculate the horizontal component of the applied force and the net force.
The horizontal component is still given by 300 N * cos(30 degrees) = 259.8 N.
The net force can be calculated using Newton's second law, F = M * a, where M is the mass of the crate (6.00 kg) and a is the acceleration (1.50 m/s^2). So, the net force is 6.00 kg * 1.50 m/s^2 = 9.00 N.
Setting up the equation for the net force:
259.8 N - 208.8 N * f = 9.00 N.
Solving for f, we get:
f = (259.8 N - 9.00 N) / 208.8 N ≈ 1.229.
Therefore, the coefficient of friction present if the crate accelerates at 1.50 m/s^2 is approximately 1.229.