# When is the direction of the static friction negative?

i thought that the force of static friction exerted on an object is always going in the opposite direction of any other force exerted on the same object. however, this problem seems to disregard that fact

The coefficient of static friction between your coffee cup and the horizontal dashboard of your car is μ= 0.800. (a) How fast can you drive on a horizontal roadway around a right turn of radius 30.0 m before the cup starts to slide?

to solve this problem, we need to find the centripetal acceleration and then use Newton's second and third laws. i.e if we say that

$F[c] = \frac{mv^2}r$ centripetal force

that means that the car have to give back that same force in the opposite direction (third law) and that is the force that will cause the cup to start sliding. so to actually measure the speed it takes to move the cup, I said that

$-F[s] = F[c]$

where the negative indicates that the force of friction is in the opposite direction. so

$$-μ(n) = \frac{m*v^2}r => -μ(mg) = \frac{mv^2}r => -μ(g) = v^2/r => v = \sqrt{-μ(g)(r)}$$

this answer doesn't make sense because of the negative under the square root. and the only way to solve this problem, is by saying that F[s] = F[c] (i.e. get rid of the negative) which does not make any sense to me. can somebody please explain to me why should the negative be neglected in this problem. thanks

• Welcome to Physics Stack Exchange. Instead of using code blocks to write equations, you can use the super awesome mathjax system. Please edit the post to use mathjax. – DanielSank Feb 23 '16 at 9:07

You write:

$$-F_s = F_c$$

and your reasoning is that if you have a centripetal force $F_c$ inwards then there must be a static force $F_s$ outwards that resists this force. The trouble is that you are starting with the observation that the cup is stationary with respect to the car, but the car is a non-inertial reference frame and Newton's laws do not apply to it. While physicists sometimes use non-inertial frames they are full of traps for the unwary and are best avoided if at all possible.

In this case, suppose you are standing by the road watching as the car races past. Newton's first law tell us that if no net force acts on an object then it will move in a straight line at a constant speed. So if the net force on the cup was zero the cup would move in a straight line and fly out of the car window.

What we actually observe is that the cup is moving in a curve, so there must be a non-zero force acting on it. The only thing acting on the cup is the dashboard surface, so the dashboard must be applying the force that is making the cup accelerate, and the only force between the dashboard and the cup is the frictional force $F_s$.

So the point is that the frictional force is the centripetal force, that is:

$$F_s = F_c$$

That's why the centripetal and frictional forces have the same sign, because they are the same thing.

• that kinda makes sense. can you explain why the cup slides in a direction opposite to the acceleration (I imagine), and not at the same direction as the car's velocity – GamefanA Feb 23 '16 at 9:11
• @user3769877: the cup just wants to go in a straight line. The cup isn't sliding away from the car, the car is accelerating away from the cup. The friction means the car can pull the cup with it as long as it doesn't accelerate too fast. The car is accelerating sideways (towards the centre of the curve) so the motion of the car relative to the cup is sideways. – John Rennie Feb 23 '16 at 9:22
• wow that makes so much sense thank you so much. but just to make sure, my problem was that i was viewing the cup from a non-inertial reference frame in which the dash is at rest and the cup is the thing in motion, when it should have been the the other way around (i.e. the dash is the thing in motion and the cup is not moving due to the friction that's acting on it) correct? – GamefanA Feb 23 '16 at 9:39
• @user3769877: correct. You have to be very, very careful when working in non-inertial frames because you need to account for all the fictional forces. Even experienced physicsts (including me!) get caught out. – John Rennie Feb 23 '16 at 9:45

You can be max when centripetal is equal to static friction ie $v^2/r=\mu g$ note it isnt $-$ as summation of external forces is zero thus we can take modulus and then equate the two forces and get the velocity.

Are you considering the problem from an inertial frame (where the centripetal force is easy to see), or a frame where the dashboard is at rest?

In the frame where the dashboard is at rest, you can simply assume that a centrifugal acceleration appears that is in the opposite direction but equal magnitude to the centripetal acceleration in the inertial frame.

that means that the car have to give back that same force in the opposite direction (third law) and that is the force that will cause the cup to start sliding.

No. The road provides a centripetal acceleration on the car, and the car provides a centripetal acceleration to the mug. The force on the mug is centripetal (in the inertial frame). It is when the acceleration is too great and the frictional force cannot provide enough to keep up with the dash that the mug begins sliding.

Whatever the type of frictional force, the frictional force will always try and reduce or prevent relative motion.

In your case if there was no friction as the car turns the corner the cup would carry on moving in a straight line.
Relative to the car this would be as a movement towards the side of the car furthest away from the centre of the corner.

So applying the rule mentioned above the frictional force on the cup would act towards the centre of the corner.