Frictional Force confusion

1. Oct 1, 2014

Redfire66

Actually I've been thinking about friction lately and it got me a bit confused
I was always taught that friction is in the direction that opposes motion, however whenever I learned about static friction I sort of just went with it. It made sense, but I never actually thought about vectors (which seemed pretty silly now that I look back)
Let's use a common example and say that a block is inclined on a ramp with an angle. If it were to be tied to something, there would be a tension force as the block comes to rest.
Is the static friction now acting on the block opposing gravity or is it opposing the tension force? Correct me if I'm wrong but I assume that it would oppose the gravitational force as gravity still acts on it... If I didn't explain clear enough then I don't mind being asked to clarify it some more
Thanks

Last edited: Oct 1, 2014
2. Oct 1, 2014

Staff: Mentor

These problems are easiest if you draw a free body diagram with all the forces acting on the block.

The block isn't moving, so you know the vector sum of all the forces will be zero. You have the tension in the rope acting on the block, gravity pulling the block downwards (decompose it into two vectors, one parallel to the ramp and one perpendicular to it), the normal force the ramp exerts on the block... and the static frictional force will be whatever is needed to make it all come out to zero. It may be acting in the same direction as the tension in the rope or in the opposite direction, depending on how great the tension is.

The standard equation for friction force $F_f=k_fF_n$ where $k_f$ is the coefficient of static friction can be misleading. It doesn't doesn't tell you what the frictional force is, it tells you what the maximum value of the frictional force can be. If that's not enough to balance al the other forces, then the object will start moving - static friction isn't strong enough to hold it in place.

3. Oct 2, 2014

CWatters

What Nugatory said. In particular...

"The block isn't moving, so you know the vector sum of all the forces will be zero.... and the static frictional force will be whatever is needed to make it all come out to zero.... It may be acting in the same direction as the tension in the rope or in the opposite direction, depending on how great the tension is."

However it is also possible for the tension to be zero. If friction was very high the block may not slide at all when placed on the inclined surface so block may not slide down and the tension might be zero and the rope slack. In that case friction will be acting up the slope. The vector sum of all the forces will still be zero.

4. Oct 2, 2014

A.T.

Imagine the block has no friction, and is held in place by you instead. Which direction will you have to apply a force if the rope tension is zero, and which direction if the rope is pulled with much more force than needed to overcome the gravity of the block.

5. Oct 2, 2014

chaitanya

y
the force will be applied in the direction antiparallel to the component of gravitational force along the incline in the first case as seen by this equation F + T - mg sin(x)=0 where T is tension , x is the angle of incline in the second case the the value of F given by the equation will be negative hence the direction will be parallel to the component of gravitational force along the incline I.e. Against the rope tension

6. Oct 2, 2014

Redfire66

Sorry, maybe I explained it poorly, I simply wanted to ask if both values (the gravitational force and tension force) were unknown, how would you go about it? Since you don't know the values of this, how would you propose to determine the value of static friction as either positive or negative in any standpoint
For example
Fnet = Fg - Ft +/- Fs where Fs is the static friction. I'm not solving any equations, just wondering about these things theoretically

7. Oct 3, 2014

CWatters

It's not as simple as Fnet = Fg - Ft +/- Fs because these are vectors that don't point in the same direction.

First you would draw a free body diagram.

Then since the block isn't accelerating in any direction you would recognise that the net force in any direction must sum to zero. To simplify the sum you could pick a direction such as parallel to the slope. Note that "parallel to the slope is parallel to Friction and Tension but NOT gravity.

Then you decide which direction is +ve. Lets say up the slope is +ve.

Then you sum(add) the components taking into account the direction. If you wanted to write it out in full it would be..

+ (+)Ft + (?)Fs + (-)mgSin(theta) = 0

but most people would skip straight to..

Ft + Fs - mgSin(theta) = 0

The - sign is because because gravity acts down the slope. Ft is +ve because it points up the slope. We don't know the sign of Fs yet. Then you solve for Fs and substitute values..

Fs = mgSin(theta) - Ft

If Fs comes out -ve then friction is acting down the slope. If it comes out +ve it's acting up the slope. Job done.

Note that the problem has two unknowns. Ft and Fs. I don't believe you can determine how much of the blocks weight is supported by friction vs tension unless you have more information about one of them (eg the tension).

Edit: I've edited this post to change Cos(theta) to Sin(theta)

8. Oct 3, 2014

Redfire66

I simply assumed in the horizontal where i didnt include any trigonometric identities as this is just theoretical. But i suppose it isn't possible to find this frictional force as you said due to lack of information. Maybe i should just research since more.
Anyway thanks for the discussion

Last edited: Oct 3, 2014
9. Oct 3, 2014

Staff: Mentor

You can't. You'll end up with one equation and two unknowns so there will be an entire family of possible solutions.