Acceleration with force of gravity and friction

[djokoman95]

Homework Statement

A block is placed against the vertical front of a cart. The coefficient of static frction between the block and the cart is [itex]\mu[/itex]_s. What acceleration must the cart have in order that the block does not fall? How would an observer on the cart describe the behavior of the block?

Homework Equations

F=[itex]\mu[/itex]_s*n
F=ma

The Attempt at a Solution

I have no idea how to start this problem. I know that the force pushing up on the block is equal to mg. I tried drawing acceleration as a horizontal vector, then setting a=mg, then trying to solve, but that didn't quite work out..

 

[Doc Al]

I know that the force pushing up on the block is equal to mg.

Good. What other forces act on the block?

I tried drawing acceleration as a horizontal vector, then setting a=mg, then trying to solve, but that didn't quite work out..

'mg' is a force, which can't be set equal to an acceleration.

Identify all the forces and apply Newton's 2nd law.

 

[djokoman95]

so after trying a few things, i got[itex]\mu[/itex]_s /g for my answer. I think that should be right.

 

[Doc Al]

so after trying a few things, i got[itex]\mu[/itex]_s /g for my answer. I think that should be right.

Almost.

 

[rwisz]

I would approach this problem by first understanding the forces at play. The block is being held against the vertical front of the cart by the force of static friction, which is equal to the coefficient of static friction (\mus) multiplied by the normal force (n) exerted by the cart on the block. The block also experiences the force of gravity (mg) pulling it down.

To prevent the block from falling, the cart must have an acceleration that is equal and opposite to the acceleration due to gravity (g). This means that the net force on the block must be zero, as stated by Newton's Second Law (F=ma).

Therefore, the equation for this problem would be:

F(net) = F(friction) + F(gravity) = 0

Substituting in the equations for friction and gravity, we get:

F(net) = \mus*n + mg = 0

Rearranging for the normal force, we get:

n = -mg/\mus

This means that the normal force must be equal to the weight of the block (mg) multiplied by the inverse of the coefficient of static friction (-1/\mus).

Now, an observer on the cart would see the block staying in place and not falling, as the block is not experiencing any acceleration. They would also notice that the block is being held in place by the force of static friction, which is equal to the normal force multiplied by the coefficient of static friction. So, they would describe the behavior of the block as being in a state of equilibrium, where the net force on the block is zero.