How Do Static and Kinetic Friction Affect Crate Movement?

[migangel]

I have a question that I am having trouble with hopefully someone can help me. Here it is
Static and Kinetic Friction:

A 10 kg crate is at rest on a horizontal table (s =0.8, k =0.3). Then various amounts of horizontal pushing force are applied to it.
a.) The largest pushing force that can be applied before the crate starts moving is of size:

N

b.)-d.) If the pushing force were instead the two amounts shown below, please fill in the rest of the values on the chart below:

horizontal pushing force: friction force: (size only)
(indicate static or kinetic):
net force on crate:
(size only)
acceleration of crate:
(size only)
13 N N, ---Select--- static kinetic N m/s2
95 N N, ---Select--- static kinetic N m/s2

 

[jbsteele]

a.) The largest pushing force that can be applied before the crate starts moving is 80 N (8 kg x 10 m/s^2).b.) If the pushing force is 13 N, the friction force is 8 N (static friction) and the net force on the crate is 5 N (13 N - 8 N). The acceleration of the crate is 0 m/s2 (as it is at rest).c.) If the pushing force is 95 N, the friction force is 30 N (kinetic friction) and the net force on the crate is 65 N (95 N - 30 N). The acceleration of the crate is 6.5 m/s2 (65 N / 10 kg).d.) If the pushing force is greater than 95 N, the friction force will remain at 30 N (kinetic friction) and the net force on the crate will be equal to the pushing force minus 30 N. The acceleration of the crate will be equal to the net force divided by the mass of the crate, (e.g. pushing force of 105 N minus 30 N friction force is 75 N, so the acceleration will be 7.5 m/s2 (75 N / 10 kg)).

 

[baba89]

I would approach this question by first understanding the definitions of static and kinetic friction. Static friction is the force that keeps an object at rest on a surface, while kinetic friction is the force that opposes the motion of an object on a surface.

a.) To determine the largest pushing force that can be applied before the crate starts moving, we need to find the maximum static friction force. This can be calculated by multiplying the coefficient of static friction (µs) with the normal force (N) of the crate. In this case, the normal force is equal to the weight of the crate (mg). Therefore, the maximum static friction force is 0.8 x (10 kg x 9.8 m/s^2) = 78.4 N. This means that the largest pushing force that can be applied before the crate starts moving is 78.4 N.

b.) For the first case where the pushing force is 13 N, the friction force will be equal to the pushing force as it is less than the maximum static friction force of 78.4 N. So, the friction force will be 13 N. Since the crate is still at rest, the net force on the crate is 0 N, and the acceleration of the crate is also 0 m/s^2.

For the second case where the pushing force is 95 N, the friction force will now be equal to the maximum static friction force of 78.4 N. This means that the crate will start moving, and the friction force will now be kinetic friction. The net force on the crate will be the difference between the pushing force and the kinetic friction force, which is 95 N - 78.4 N = 16.6 N. The acceleration of the crate can be calculated using Newton's second law, F=ma, where F is the net force and m is the mass of the crate. Therefore, the acceleration of the crate will be 16.6 N / 10 kg = 1.66 m/s^2.

In conclusion, the values for the chart would be:
horizontal pushing force: 13 N, 95 N
friction force: 13 N, 78.4 N
static or kinetic: static, kinetic
net force on crate: 0 N, 16.6 N
acceleration of crate: 0 m/s^2, 1.66 m/s^