Solving Kinetic Friction: Min Force Needed to Prevent Sliding

[devnull]

5. The drawing shows a large cube (mass = 23 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass = 3.2 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of static friction between the cubes is 0.71. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?


I'm having trouble writing a force diagram. I know the formulas to use but I don't know which direction the forces are in. Any help would be appreciated.

 

[Doc Al]

Static friction, not kinetic

Originally posted by devnull
I'm having trouble writing a force diagram. I know the formulas to use but I don't know which direction the forces are in.

Give it a shot. Take each cube and write down all the forces acting on it. (Don't forget Newton's 3rd law.)

Hint: Besides the applied force P, I see three "kinds" of forces: weight, friction, and normal forces.

 

[M98Ranger]

Sure, I can try to help you with the force diagram. Let's start by labeling the forces involved in this scenario. We have the force P acting on the large cube, the weight of the large cube (mg), the normal force from the surface on the large cube (N), and the force of static friction between the two cubes (fs).

Now, let's consider the motion of the small cube. Since it is in contact with the front surface of the large cube, it will experience the same acceleration as the large cube. Therefore, the only force acting on the small cube is the force of static friction (fs) from the large cube.

Next, let's consider the motion of the large cube. Since it is being accelerated by the force P, the net force on the large cube is P - fs. In order to prevent the small cube from sliding downward, the force of static friction (fs) must be equal to or greater than the weight of the small cube (mg). This can be represented as fs ≥ mg.

Now, let's use the formula for static friction, which is fs = μsN, where μs is the coefficient of static friction and N is the normal force. We can rearrange this formula to solve for N, which gives us N = fs/μs.

Substituting this into our previous inequality, we get fs ≥ mg = (3.2 kg)(9.8 m/s^2) = 31.36 N.

Now, we can substitute this value for fs in our net force equation: P - 31.36 N ≥ 0. This means that the minimum magnitude for P is 31.36 N in order to prevent the small cube from sliding downward.

I hope this helps with your force diagram and understanding of the problem. Let me know if you have any further questions.