# Friction and slope

Problem:

A body lays still on a sloping plane with slope angle alpha. How high does the friction coefficient need to be in order to allow for this equlibrium?

Thoughts:

I attempted solving this by basic trig. cos(a) = oposing/hypothenuse. So say the opposing is 1 unit, then the friction coefficient f_k needs to balance the vector acting in the direction of the hypotenuse downwards. So:

cos(a) = 1/f_k

f_k = 1/cos(a).

However the answer is supposed to be f_k = tan(a). How come?

## Answers and Replies

You are on the right track that the "friction" needs to balance the "vector" acting along the hypotenuse. What you are missing is that the friction is a function of the normal force as well as the friction coefficient.

Start by drawing a free body diagram of the body on the slope.

You are on the right track that the "friction" needs to balance the "vector" acting along the hypotenuse. What you are missing is that the friction is a function of the normal force as well as the friction coefficient.

Start by drawing a free body diagram of the body on the slope.
Hi, thanks for your reply. I'm not sure how I should proceed after drawing the free body diagram? Obviously the body is affected by vertical gravitational pull $$F=mg$$ and another force parallell to the horizontal ground plane. The resultant of these two forces is the force acting on the body in the direction of the hypotenuse?

What is "another force"?

The horizontal component of the vector parallell to the hypotenuse?

OK, but does this vector have a name?

Can you provide an attempt at drawing the free body diagram?

See attached. Calculating by it I get something like

$$\sin\alpha = \frac{F_v}{F_k} \ \Rightarrow \ \cos\alpha = \frac{F_h}{F_k}= \frac{F_h}{F_v/\sin\alpha}$$

which gives me an expression for Fh as

$$F_h = F_v\frac{\cos\alpha}{\sin\alpha} = F_v\cot\alpha.$$

Setting the two Fk's, expressed in sin and cos respectively, equal it yeilds

$$\frac{F_v}{\sin\alpha} = \frac{F_h}{\cos\alpha} \ \Leftrightarrow \ \frac{F_v}{\sin\alpha} = \frac{F_v\frac{\cos\alpha}{\sin\alpha}}{\cos\alpha},$$

Which just teaches me that sin a = sin a, as if I didin't know that.

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BiGyElLoWhAt
Gold Member
I think you're making it way harder than it needs to be. I also think it's good to go down these dead ends every once in a while as it will help you avoid them in the future.

First that freebody diagram isn't any good. How is that block going to move based on the lengths and directions of the forces?

Not sure what you're asking? It's going to move in the direction of Fk?

BiGyElLoWhAt
Gold Member
Looking again, the motion might end up alright, but I still don't like it. I have no idea what you're calling your forces and there's a little bit of inconsistancy in there.

BiGyElLoWhAt
Gold Member
There are 2 things that are exerting a force on the block. What are they?

Gravity and friction?

BiGyElLoWhAt
Gold Member
No, those aren't things. Hint, gravity comes from the earth. Where does friction come from?

Surface? It comes when two surfaces move in relation to each other in different directions. I feel like a donkey now to be honest, I know what friction is but I'm not sure what you want me to realize is relevant to the problem.

BiGyElLoWhAt
Gold Member
Is friction the only force that comes from the surface?

Or,
What is the equation for friction?

BiGyElLoWhAt
Gold Member
And don't feel bad. You're overlooking something simple, and overcomplicating the problem. I'm not trying to make you feel stupid, just realize what you overlooked =]

Is friction the only force that comes from the surface?

Or,
What is the equation for friction?
What about the normal force perpendicular to the surface? But I'm not sure how to draw out the relevant forces on the free body diagram?

BiGyElLoWhAt
Gold Member
Yea the normal is what I was getting at. You have a normal Force, a frictional force, and a gravitational force. You want the system to be in equilibrium, so what does this mean mathematically?

In other words, give me an example of an equation of a system that's in equilibrium.