Minimum coefficience of friction PLEASE HELP

[jtak77]

Homework Statement

A box with a mass of 20N is tied to another box with a mass of 15N hung over a pulley on an incline of 10 deg. please find the minimum coefficience (mu) for remaining stationary

Homework Equations

Not too sure, F(mu)=(mu)N N being normal force

The Attempt at a Solution

3.47/19.70 = (mu)19.70/19.70

mu =0.176

Im not sure if there are more steps beyond this, but this is what I got for mu any help on this please?

 

[anathema]

Hello,

To find the minimum coefficient of friction (mu) for the system to remain stationary, we need to consider the forces acting on the boxes. These include the weight of the boxes, the normal force, and the frictional force.

The weight of the boxes can be calculated as follows:

Box 1: m1 = 20N
Box 2: m2 = 15N

The weight of each box can be calculated using the formula W = mg, where g is the acceleration due to gravity (9.8 m/s^2).

Box 1: W1 = 20N * 9.8 m/s^2 = 196N
Box 2: W2 = 15N * 9.8 m/s^2 = 147N

Next, we need to consider the normal force. This is the force exerted by the surface on the boxes in a direction perpendicular to the surface. In this case, the normal force is equal to the weight of the boxes since they are resting on a flat surface.

Therefore, the normal force for both boxes is equal to their respective weights: N1 = W1 = 196N and N2 = W2 = 147N.

Now, we can calculate the frictional force using the formula F = mu*N, where mu is the coefficient of friction and N is the normal force.

For the system to remain stationary, the frictional force must be equal to or greater than the force pulling the boxes down the incline (the weight component along the incline). This can be represented by the equation:

Ffriction = mu*N = mg*sin(theta)

Where theta is the angle of inclination (10 degrees in this case).

Substituting the values we calculated earlier, we get:

mu*196N = (20N + 15N)*sin(10 deg)
mu = (35N*sin(10 deg))/196N
mu = 0.096

Therefore, the minimum coefficient of friction for the system to remain stationary is 0.096.

I hope this helps! Let me know if you have any further questions.

 

[tomcue]

I would like to clarify that there is no such thing as a "minimum coefficient of friction." The coefficient of friction is a constant value that depends on the materials in contact and the surface conditions. It is not something that can have a minimum value.

In this scenario, the coefficient of friction is not needed to determine the minimum force required to keep the boxes stationary on the incline. The only forces acting on the boxes are gravity and tension in the rope. The minimum force required to keep the boxes stationary is equal to the weight of the boxes, which is 35N.

If you are trying to find the coefficient of friction between the boxes and the incline, you would need to use the equation F(friction)=μN, where N is the normal force exerted by the incline on the boxes. This normal force can be calculated by breaking down the weight of the boxes into components parallel and perpendicular to the incline.

However, it is important to note that friction is not the only force acting on the boxes in this scenario. The tension in the rope also plays a role in keeping the boxes stationary. Therefore, the calculation of the coefficient of friction in this situation may not accurately represent the true frictional forces at play.