# Force on an object placed on an incline

1. Sep 20, 2014

### reigner617

The Question: An ice block of mass M=80.0 kg is held stationary on a frictionless ramp. The ramp is at an angle of 36.9 degrees above the horizontal. If the block is held in place by a horizontal force directed horizontally toward the center of the block, find the magnitude of this force.

Related Equations: F = mgsinθ, F = mgcosθ

The Solution: Our professor solved it as follows:
Fcosθ = mgsinθ
F= mgsinθ/cosθ
= 470.7/ cos 36.9
= 588.6 N

My only confusion is to why he took this approach. I understand that when the force is tangential, we use mgsinθ; when the force is perpendicular, we use mgcosθ, but there doesn't seem to be a "formal" formula for a horizontal force

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2. Sep 20, 2014

### Bandersnatch

Hi reigner617

Perhaps it'd be easier to see if you stopped thinking in terms of "formal formulas" and instead tried to draw a free body diagram and split the two forces in play(the horizontal F and the vertical weight) into component forces parallel and perpendicular to the incline.
As long as you're ok with trigonometry, you should see where all the sines and cosines come from.

3. Sep 20, 2014

### reigner617

I drew a diagram with mg directly perpendicular to the ground and the horizontal force parallel to the ground. I understand how mgsinθ and mgcosθ are derived, but I'm still having trouble understanding as to why F horizontal would equal mgsinθ/cosθ.

4. Sep 20, 2014

### Satvik Pandey

I have made a figure for you.From this figure can you resolve 'F' on the green line.
Now if the blocks are stationary then what should be the net force acting on the block.

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5. Sep 20, 2014

### reigner617

I figured out what was drawn on the diagram. To answer your question, the net force has to be zero, but I just don't understand where cosθ came from. I apologize if I sound really dense right now. I'm sure I'm forgetting something very basic, but whatever that is i can't quite figure out.

6. Sep 20, 2014

### reigner617

I labeled the green line as f for friction. With that said, i could say cosθ = f/F or f=Fcosθ. I'm assuming f= mgsinθ and from that came the answer; however, taking friction into account does not seem right because the problem says that it is a "frictionless" ramp, so I understood that to be f=0 (or negligible).

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7. Sep 21, 2014

### Satvik Pandey

You need to find the component of 'F' along the green line.For doing this you need to draw a line passing from the head of vector 'F' perpendicular to the green line.
Please look at the diagram.
Can you now find projection of 'F' on the green line?

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8. Sep 21, 2014

### ehild

Satvik, upload your picture with the image button so people can see it immediately.

ehild

9. Sep 21, 2014

### Satvik Pandey

Thanks ehild next time I will use that button.:D

10. Sep 21, 2014

### reigner617

So using f for friction is correct? And if I do that and use cosθ, I get cosθ=f/F or Fcosθ=f. Am I supposed to assume that f=mgsinθ? This doesn't seem right because the ramp is frictionless...

11. Sep 21, 2014

### Bandersnatch

There is no friction. f is just a component of the force F parallel to the incline(the only one that matters here).

How well do you know you trigonometry? If you drew a right triangle ABC with right angle at vertex B and angle alpha at vertex A, would you be able to represent relationships between the lengths of the sides using trig functions of angle alpha? This is all that's going on here.

12. Sep 21, 2014

### Satvik Pandey

There is no frictional force in this question.The component of force(mg) along the incline(mgsinθ) is balanced by the component of force along the incline(Fcosθ).
So your equation becomes

mgsinθ=Fcosθ.
Find 'F' from this equation.