# Finding direction of friction

1. Feb 23, 2013

### Saitama

1. The problem statement, all variables and given/known data
A plank of mass M is placed over smooth inclined plane and a sphere is placed on the plank as shown (see attachment). There is sufficient friction between sphere and plank to prevent slipping. If system is released from rest, the frictional force on the sphere is
a)up the plane
b)down the plane
c)zero
d)horizontal

2. Relevant equations

3. The attempt at a solution
I began by drawing a FBD of the sphere (see attachment 2). I assumed that the friction,f acts down the plane.
Equation for translational motion of sphere:
$mg \sin \theta=ma$...(i)
where a is acceleration of sphere
Equation for rotational motion of sphere (taking torque about CM):
$fR=Ia/R$....(ii)
where I is the moment of inertia of sphere and R is the radius of sphere. Solving the equations, I get a positive value of f which means that the direction of friction is down the plane. But the answer is zero.

Any help is appreciated. Thanks!

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• ###### friction2.png
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Last edited: Feb 23, 2013
2. Feb 23, 2013

### Astronuc

Staff Emeritus
Friction opposes motion. If the ball/log were to slipped, what would be the direction of motion. The ball rolls because something opposes it sufficiently to prevent slipping.

3. Feb 23, 2013

### Saitama

Its direction would be opposite to the slipping. But I still don't understand why its zero during rolling?

4. Feb 23, 2013

### Staff: Mentor

You have to consider the motion of the plank as well. How does their relative acceleration look like without friction?

5. Feb 24, 2013

### Saitama

Sorry for being late.

The relative acceleration without friction is zero.
Applying Newton's second law for plank:
$$Mg\sin \theta-f=Ma'$$
The direction of frictional force is up the plane on the plank.
$$a'=\frac{1}{M}(Mg\sin \theta-f)$$
In the reference frame fixed to the plank, a pseudo force, $ma'$ acts on the sphere up the plane.
$$mg\sin \theta+f-\frac{m}{M}(Mg\sin \theta-f)=ma$$
$$f+\frac{mf}{M}=ma$$
$$f=\frac{mMa}{m+M}$$
Taking moments about the CM of sphere
$$fR=I\frac{a}{R}$$
Substituting the value of a
$$fR-f\frac{M+m}{mMR}=0$$
$$f=0$$
This gives me the right answer. Is this method correct? Is their any easier way?

Thanks mfb!

Last edited: Feb 24, 2013
6. Feb 24, 2013

### haruspex

Rather. Suppose there were no friction between the sphere and the plank. Each would experience a component of force down the plane in proportion to its own mass, so they would accelerate together. Thus, no sliding will occur between them. Hence, there is no frictional force even when the surfaces are rough.

7. Feb 24, 2013

### Saitama

I understood what you said before this but how did you reach this conclusion?

Last edited: Feb 24, 2013
8. Feb 25, 2013

### haruspex

Frictional forces only arise when, in the absence of those forces, the two surfaces would have slid in relation to each other. If there was no tendency for them to move out of synch then no frictional force.