# Force Needed to Keep Mass on Wedge Stationary

• Violet42
In summary, the conversation discusses a problem involving a wedge of mass M on a flat horizontal surface with an incline of 20 degrees and a friction coefficient of 0.020. A vertical downward force of 3N is being applied to the wedge by a mass resting on top. The question is to determine the force needed to keep the mass stationary. The conversation includes discussions of friction, resolving forces, and combining fundamental concepts with empirical laws. However, there is some confusion about the interpretation of the problem and more information is needed to fully understand the situation.
Violet42

## Homework Statement

Hi, I'm new and I've been trying to figure this problem out forever. I can't even really say I've attempted a solution because I don't really undertand the concept very well. (I'm paraphrasing the question here)
A wedge of mas M (1.00[kg]) is at rest on a flat horizontal surface. The wedge is at an incline of 20 degrees. The horizontal surface has a friction coefficient of 0.020. There is a completely vertical (downward force) being applied to the wedge by a mass resting on its top. The force is 3[N]. What force must be applied to the wedge to keep the mass stationary?

## Homework Equations

I don't even know where to begin, but question is from an 'energy' unit, so if there's some way that this is related to energy...

But I do have:
F of Friction= Friction coefficient * Normal force
Normal Force= mgsinA
Parallel Force=mgcosA

## The Attempt at a Solution

Like I said before, I'm truly lost. I have no grasp of what's going on here whatsoever, so if someone could really break this down step by step for me that would be wonderful.
I'm basically trying to identify all the forces in the system and make the netforce equal zero, right?
Thank you sooo much.
Violet

## The Attempt at a Solution

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More than everything you need to know is here http://en.wikipedia.org/wiki/Friction

Rather useful to note, e.g. for your problem, is the "angle of friction".

Your problem is partly about resolving forces into their components in a given direction which you must have done. And then it is about combining that with an odd piece of extremely phenomenological (i.e. very empirical and non-fundamental) physics - the laws of static friction. So get on top of that - it is not very difficult. Nor, I add, very interesting - an added motive to just get on top of it quickly. I guess you do often have to combine one part fundamental math/physics concepts with purely empirical laws. Sort of engineering maths [PLAIN]http://img84.imageshack.us/img84/8527/grinxf2.gif[SIZE="1"][COLOR="DarkOrange"]sneer[/COLOR][/SIZE].

Feynman was a good enough lecturer to try and make friction more interesting in his lectures, see ref 15. Also there was years ago an article about it in the old Scientific American which sometimes comes to my mind at the time of the jerk you get in the last half-second when you smoothly brake a car (that is dynamic friction though).

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I'm sorry if I'm just not understanding what you're saying, but there is no friction between the mass and the wedge, only between the wedge and the surface it's resting on.

Violet42 said:
I'm sorry if I'm just not understanding what you're saying, but there is no friction between the mass and the wedge, only between the wedge and the surface it's resting on.

So in other words, the mass is to be held in place due to the assembly being accelerated by an external force? That would mean that you want the net force exerted by the mass on the wedge to be normal to the wedge surface (i.e., no component parallel to the surface).

A sneaky way to look at it is, what acceleration vector would you have to add to that of the acceleration due to gravity in order to "tilt" the direction of "down" to be normal to the wedge slope?

Violet42 said:
I'm sorry if I'm just not understanding what you're saying, but there is no friction between the mass and the wedge, only between the wedge and the surface it's resting on.

OK, everything I said is true and maybe you know it already. On rereading more carefully the question I do not understand what situation the words describe and how it is a problem of static friction. I'd need a picture of it.

If I've interpreted the question correctly, the situation is as portrayed in the attached image.

#### Attachments

• fig10.gif
2.2 KB · Views: 493
Your diag is what I thought - I can only think we have a problem of exegesis before physics.

In your diag I can see nothing (no horizontal force) tending to make the wedge move. The weight you put on top of the wedge could slide down but somehow the wording implies this is not part of the problem, maybe the friction is high, anyway there is no information about it. If this is being marked by a human rather than a computer, the student* could answer that it will not move if any horizontal force up to so much is applied. (Other interpretations of what the question means are possible, but it seems to me ambiguously written.)

*You appear to be teacher rather than student. ?

epenguin said:
Your diag is what I thought - I can only think we have a problem of exegesis before physics.

In your diag I can see nothing (no horizontal force) tending to make the wedge move. The weight you put on top of the wedge could slide down but somehow the wording implies this is not part of the problem, maybe the friction is high, anyway there is no information about it. If this is being marked by a human rather than a computer, the student* could answer that it will not move if any horizontal force up to so much is applied. (Other interpretations of what the question means are possible, but it seems to me ambiguously written.)

*You appear to be teacher rather than student. ?

Once again, this is just my interpretation of the question wording; it could well be that I've got hold of the wrong end of the stick, so to speak.

In my diagram I indicated a force acting on the left side of the wedge. The idea is that this force should be sufficient to accelerate the wedge and block to the right at a rate such that the block remains stationary on the zero friction surface of the wedge.

Perhaps the Original Poster can give a yea or nay to this scenario.

## What is force needed to keep mass on wedge stationary?

The force needed to keep a mass on a wedge stationary is equal to the component of the weight of the mass that is parallel to the surface of the wedge. This is known as the normal force and is calculated using the formula F = mgsinθ, where m is the mass, g is the acceleration due to gravity, and θ is the angle of the wedge.

## What factors affect the force needed to keep mass on wedge stationary?

The force needed to keep a mass on a wedge stationary is affected by the mass of the object, the angle of the wedge, and the force of gravity. Other factors that may affect the force needed include friction and the surface of the wedge.

## How do I calculate the force needed to keep mass on wedge stationary?

To calculate the force needed, you can use the formula F = mgsinθ, where m is the mass, g is the acceleration due to gravity, and θ is the angle of the wedge. Simply plug in the values for each variable and solve for F.

## What happens if the force needed is not applied?

If the force needed to keep a mass on a wedge stationary is not applied, the mass will start to slide down the wedge due to the force of gravity. This is because the normal force is not strong enough to counteract the force of gravity, resulting in motion.

## How does the force needed change as the angle of the wedge increases?

The force needed to keep a mass on a wedge stationary increases as the angle of the wedge increases. This is because as the angle increases, the component of the weight of the mass that is parallel to the surface of the wedge also increases, requiring a larger normal force to keep the mass stationary.

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