Help with force and friction on a flat surface

[Mattches]

Homework Statement

I have a 30kg crate on a flat floor. There is a rope pulling the crate to the right with a force of 125N. The coefficients of static and kinetic friction are .34 and .24 respectively. I need to know what the minimum force I could put on top of the crate to stop it from moving.
Fg = 9.81 m/s^2
uS = .34
uK = .24
m = 30kg

Homework Equations

I know that I need to use Newtons second law to determine the sum of all forces in each direction, and see how much needs to be added to overcome the 125N pull. But honestly I have no idea where to start!

The Attempt at a Solution

I know that Fg on the crate is 294.3N. And I know that friction opposes the way the crate is being pulled. But I really don't know how to piece this problem together.
Can anyone help?

 

[Doc Al]

What must the total static friction be to keep the crate from moving? (Assuming it is initially at rest.)

How is max static friction related to the normal force between crate and floor?

 

[Mattches]

The total static friction has to be equal to the force pulling the crate. The normal force is pushing the crate up at the same rate that gravity is pulling down but...I'm still not exactly sure where to go from there. Sorry but I'm very new at this :[

 

[Doc Al]

The total static friction has to be equal to the force pulling the crate.

Good.

The normal force is pushing the crate up at the same rate that gravity is pulling down but...

That would be true if you didn't apply any additional force on the crate. But whatever force you push down on the crate with will add to the normal force.

How does static friction relate to the normal force? Use that fact (and your first statement in this post) to figure out what the total normal force needs to be for there to be enough friction.

 

[rwisz]

As a scientist, it is important to approach problems in a systematic and logical manner. In this scenario, the first step would be to draw a free body diagram of the crate, showing all the forces acting on it. This will help in visualizing the problem and identifying the key components.

Given the mass of the crate (30kg), the force of gravity acting on it can be calculated using the formula Fg = mg, where m is the mass and g is the acceleration due to gravity (9.81 m/s^2). In this case, Fg = 294.3N.

Next, we need to consider the force of friction acting on the crate. Friction is a force that opposes motion, and it can be either static or kinetic depending on the situation. In this case, the crate is not moving initially, so we can assume that the force of friction is static. The formula for static friction is Fs = uS * N, where uS is the coefficient of static friction and N is the normal force (the force exerted by the surface on the object). In this scenario, N is equal to the force of gravity, so Fs = uS * Fg = 0.34 * 294.3 = 99.9N.

Since the crate is being pulled to the right with a force of 125N, we can determine the net force acting on the crate by subtracting the force of friction from the pulling force. Net force = 125N - 99.9N = 25.1N.

According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is 25.1N and the mass is 30kg, so we can calculate the acceleration of the crate using the formula a = Fnet/m = 25.1N/30kg = 0.83 m/s^2.

Now, to determine the minimum force needed to stop the crate from moving, we need to consider the force required to counteract this acceleration. This force can be calculated using the formula F = ma, where m is the mass and a is the acceleration. In this scenario, F = 30kg * 0.83 m/s^2 = 24.9N.

Therefore, the minimum force that needs to be applied on top of the crate to stop it from moving is approximately 24.