# Accelerating wedge at angle above the horizontal with friction

1. Nov 13, 2012

### PhoniexGuy

1. The problem statement, all variables and given/known data

http://imageshack.us/a/img594/4821/fileeq.jpg [Broken]

So, the problem is to find the minimum acceleration required to keep the block on the ramp, when the ramp is being pushed at an angle to the horizontal (so going diagonally)

I know that the coefficient of static friction is 0.4 and gravity is 10.0m/s/s

2. Relevant equations

I think these are what i need to use:

Fnet = ma
FfStatic = uFn
And maybe need w = mg?

3. The attempt at a solution

Well, honestly i have no idea what to do, the final answer is 3.84, but i don't get how to get it.

I tried to do something like this: Fn = 10*sin(30)/0.4 = 12.5, and then the friction force is 0.4*12.5 = 5.0, but no idea what that means. Any help?

Last edited by a moderator: May 6, 2017
2. Nov 13, 2012

### haruspex

What are all the forces on the block? You have mg, Fn and uFn, right? In what direction is the net force on the block? What equation can you write down for the forces perpendicular to the net force?

3. Nov 13, 2012

### PhoniexGuy

i don't have the weight of the block, the problem doesn't state that. Is it in the direction the applied force is pushing? I honestly don't know...

4. Nov 13, 2012

### PeterO

I am not surprised.

The problem refers to the minimum acceleration on the block.

I have only ever seen acceleration used with of, not on - such as what is the minimum acceleration of the block. Is that what is meant here?

It is also a pity that the diagram is so poor. That 30 degree angle is drawn as almost 45, so the visual feedback you are getting is not overly helpful.

With a 30 degree ramp, and a force applied at 60 degrees to the horizontal, it actually means the force, and thus acceleration, is perpendicular to the contact surface.

Now to this problem.

If the block and ramp were stationary you can calculate the component of the weight [mg] acting down the slope, and the component perpendicular to the slope [Fn].
Than enables you to calculate the friction [μ.Fn]

You will see that the friction force is less than the weight component down the slope - so the block would most certainly slip down the ramp.

You could then calculate how much additional friction - achieved by having a larger Normal Force, Fn, - is needed in order for the block not to slip.

Finally, every one of those forces will have an m factor in them - since we are not told the mass of the block - and by reference to newton's Second law F = ma, we might see that of the extra force needed was 3.84m Newtons, then that could be achieved by having an acceleration of 3.84 ms-2, which was their answer. You just have to check whether your calculations yield that figure of 3.84.

Note: I have "bolded" each reference to m as mass so that it can be distinguished from the m as the unit, metres.

Last edited by a moderator: May 6, 2017
5. Nov 13, 2012

### PhoniexGuy

But thats what i don't get, how do i get the normal force, and the weight and stuff if i only have gravity and nothing else?

6. Nov 13, 2012

### PeterO

You have geometry!! the ramp is at 30o so one of them will be mgsin30o and the other will be mgcos30o

You now have to work out which one is which.

EDIT: and on the way through you will not have just numbers, you will always have a number multiplied by m

7. Nov 13, 2012

### PhoniexGuy

So it like Fn = m*g*sin(30)? Why are you using sin 30.

8. Nov 13, 2012

### PeterO

I didn't use sin 30; you did.

I said use sin30 and cos30 but that you would have to work out which one to use for which Force.

http://www.physicsclassroom.com/Class/vectors/U3L3e.cfm

9. Nov 13, 2012

### PhoniexGuy

Wow, thanks that makes alot more sense.

10. Nov 13, 2012

### PhoniexGuy

Okay, so I got the normal force is Fn = Fg*cos(30) = 8.66N, and I know that F|| = Fg * sin(30) = 5.00 N, how do I get the acceleration now? Since if I do, 0.4*Fn = 3.4 is wrong, but I need 3.84, why am I off?

11. Nov 13, 2012

### PeterO

The figures match up if the mass is 1 kg - but I won't worry about that detail at the moment.

you have that the friction force is only 3.464 N, whereas the component of weight down the slope is 5 N. [you should not have rounded off the friction to 3.4]

How much bigger would Fn have to be in order for the friction to be 5N ?

12. Nov 13, 2012

### PhoniexGuy

Needs to be 12.5 since 5/.4 = 12.5, then 12.5 - 8.66 = 3.84! Yay, thanks for the help mate!