How Does Friction Affect a Small Block on a Moving Incline?

[countryapple]

Homework Statement

A small block of mass .5kg sits on a large block of mass 5kg as shown. The coefficient of static friction between the blocks is .6, but between the large block and the table there is no friction. What horizontal force will keep the small block from slipping down the incline if the angle is 50 degrees.

Homework Equations

How do I take into account the incline is also moving?

The Attempt at a Solution

I set up a Ff< mew(sum of forces)

I solved for F, but did not get the right answer which should be around 2

 

[rl.bhat]

Hi countryapple, welcome to PF.
The problem is not clear. You have mentioned the table and inclined plane. On which the larger block is moving? To which the horizontal force is applied. If you can provide the figure the problem will be clear.

 

[kerimek]

.5N.

As a scientist, it is important to carefully consider all factors in a problem before attempting to solve it. In this case, the given situation involves a moving incline, which adds another variable to the problem. In order to take this into account, we need to use the concept of relative motion. This means that we must analyze the forces acting on each object individually, as well as the forces acting on the system as a whole.

First, let us consider the small block. The only forces acting on it are its weight, which is directed downwards, and the frictional force, which is directed upwards along the incline. Since the block is not slipping, the net force on it must be zero. Therefore, we can set up the following equation:

Ff = mg*sin(50)

Where Ff is the frictional force, m is the mass of the block, and g is the acceleration due to gravity.

Next, let us consider the large block. It is important to note that the large block is not accelerating, as there is no friction between it and the table. Therefore, the net force on the large block must also be zero. This means that the force of friction between the two blocks must be equal and opposite to the force of the small block on the large block. We can set up the following equation:

Ff = (m1+m2)*a

Where m1 and m2 are the masses of the small and large blocks respectively, and a is the acceleration of the system.

Now, we can combine these equations to solve for the horizontal force, F, that is needed to keep the small block from slipping down the incline. Setting the two equations for Ff equal to each other, we get:

mg*sin(50) = (m1+m2)*a

Solving for a, we get:

a = mg*sin(50) / (m1+m2)

Finally, we can plug in the given values (m1=0.5kg, m2=5kg, g=9.8m/s^2, and the angle of 50 degrees) to find the required horizontal force, F:

F = (0.5kg+5kg)*9.8m/s^2*sin(50) / (0.5kg+5kg)

F = 2.46N

This is close to the expected answer of 2.5N, and any small