What is the Minimum Horizontal Force Needed for a Block to Slide Up an Incline?

[anikmartin]

Greetings fellow physicists!
I am stuck on a problem, please help.
The situation is a block on a rough incline (trainglular block), the incline itself is on a frictionless table. A horizontal force is applied to the incline. The question is to find the minimum horizontal force needed so that the block starts to slide up the incline. This means the only friction force involved in this problem is static friction, and because the block will move up the incline this frictional force acts in the direction down the incline. When the horizontal force is applied to the incline the contact force between these two objects will increase, which is the normal force on the block, and thus the static friction force will increase until it reaches the maximum, and then the block will begin to move. Is this correct?

I am having trouble finding a way for the block to move up the incline. There must be a force acting on the block in this direction, but what is it? The normal force is perpendicular to the block, its weight has a force component down the incline, and static friction acts down the incline (as the horizontal froce is being applied). Is this correct? I can't seem to find the force!


Please help!

 

[Doc Al]

Welcome to PF!

When the horizontal force is applied to the incline the contact force between these two objects will increase, which is the normal force on the block, and thus the static friction force will increase until it reaches the maximum, and then the block will begin to move. Is this correct?

Yes.

I am having trouble finding a way for the block to move up the incline. There must be a force acting on the block in this direction, but what is it? The normal force is perpendicular to the block, its weight has a force component down the incline, and static friction acts down the incline (as the horizontal froce is being applied). Is this correct? I can't seem to find the force!

The force acting to accelerate the block upwards is the normal force, which certainly has a vertical component.

While it is certainly possible to view things relative to the incline, I advise against it: realize that the incline will be accelerating and thus is no longer an inertial frame of reference. Look at things relative to the table.

Now consider vertical and horizontal forces acting on the block. Just before the block starts to slide up the incline it is in vertical equilibrium with static friction at a maximum. Write the equation for the vertical forces.

Of course, horizontally the entire setup is accelerating. Apply Newton's 2nd law to the block to find its acceleration. Then figure out what the force on the incline must be to produce that acceleration.

 

[anikmartin]

inertial and noninertial frames of reference

Oh, I see, relative to the table the block has a vertical acceleration and a horizontal acceleration. Thank you! When I stepped away from the incline and onto the table, as I pictured it in my head, I could see the block moving up the inline, vertical relative to me, and at the same time horizontally with the incline, relative to me. Instead of having the x-axis parallel to the incline, I set it parallel to the table, which worked. But it is still unclear why I can't find its motion relative to the incline, the situation gets a little fuzzy when I try to think it through. Maybe the better question would be to ask: what would have to be done if I wanted to find the acceleration of the block relative to the incline? Hmmmm . . .

It is true that if the block and the incline are considered as one object then the net force on the system is F=(M+m)a, correct?

I have been reading about inertial and noninertial frames, I understand the difference, but when it comes to the application I get confused.
The rule is that as long as you are measuring <from> an inertial frame you can apply Newton's Laws, correct?

Here is a question, regarding this problem. Say there are two blocks of different masses (m1,m2) resting on a frictionless table, each in contact with each other. If a force is applied to m1, m1 and m2 will both have the same acceleration, and will move as a system. My question is regarding the contact forces between each block. My teacher said that the force from m1 on m2 will equal the force on m1 from m2, when the system is accelerating, by Newton's Third Law. Wouldn't the system have to be at rest, or have a constant velocity for this to be true?

Or is that only true if I am measuring these contact forces from, say within one of the blocks. If I was measuring the contact forces from within the blocks, they would not be equal? How would this situation be set up mathematically. I am trying to get a clear picture and understanding of how to move back and forth between noninertial and inertial frames. Okay, thanks for your help.

 

[Doc Al]

It is true that if the block and the incline are considered as one object then the net force on the system is F=(M+m)a, correct?

Right. That's true until the block starts to slide up the incline.

I have been reading about inertial and noninertial frames, I understand the difference, but when it comes to the application I get confused.
The rule is that as long as you are measuring <from> an inertial frame you can apply Newton's Laws, correct?

Right. If you do things from an accelerating frame, you'll have to add terms to Newton's laws.

My question is regarding the contact forces between each block. My teacher said that the force from m1 on m2 will equal the force on m1 from m2, when the system is accelerating, by Newton's Third Law. Wouldn't the system have to be at rest, or have a constant velocity for this to be true?

Contact forces always satisfy Newton's 3rd law, regardless of motion.

 

[aekanshchumber]

Calculate the psuedo force on the block i.e force that is opposite to the acceleration due to the forward pull. Divide the psuedo foce vector in components along and perpendiculer to the plane of the block. The component along the plane will push the block upward. easy isn't it.

 

[anikmartin]

Thank you both.