# Sliding and rolling friction

I heard that friction depends only on the normal force but not on the contact area. Let's take a cube and a sphere which are of same weight (then normal force will also same ) but the force needed to move these two objects is different, why?

• Are you assuming that the sphere won't roll? – Daniel Griscom Nov 19 '15 at 3:10
• Have a look at this: physics.stackexchange.com/q/218468. A sphere will also rotate (assuming there is enough friction), that requires extra energy, which is not needed for pure sliding. – Gert Nov 19 '15 at 3:10
• I'd like to answer this question comprehensively, assuming it doesn't get closed or assigned as 'HW&E'. I'll wait till tomorrow. – Gert Nov 19 '15 at 3:30

Let's look at this problem from the point of view of equations of motion, see diagram below:

Firstly let's make a few assumptions.

1. Ball and cube are of same weight ($mg$) and same size.
2. Simple friction model $F_f=\mu F_n$ holds and $\mu$ is independent of speed.
3. Both objects are completely stationary (no sliding, rolling or tumbling) at $t=0$, at which point a purely horizontal force $F$ starts acting on the centre of gravity of the objects.

A) Case of the Ball:

If we assume there is enough friction the ball will start translating and rolling (but without sliding or slipping) and at each instant:

$$v=\omega R,$$

with $v$ the translational speed, $\omega$ the angular speed and $R$ the radius of the ball.

For the translational speed we can set up the Newtonian equation:

$$ma=F-F_f,$$

$$ma=F-\mu mg,$$

$$m\frac{dv}{dt}=F-\mu mg.$$

Integrated between $0,0$ and $t,v$ we get:

$$v=\bigg (\frac{F}{m}-\mu g \bigg )t.$$

And:

$$\omega=\frac{v}{R}.$$

The kinetic energy $K$ of the object is now the sum of translational energy and rotational energy and also the work $W$ done on the object:

$$K=\frac{mv^2}{2}+\frac{I \omega^2}{2}=W.$$

With $I=\frac{2mR^2}{5}$ the inertial moment of the ball and inserting we get:

$$K=\frac{7mv^2}{10}=W_A.$$

B) Case of the Cube:

Although $F$ causes a tumbling moment $FR$ around the forward contact point, we'll assume no tumbling actually occurs (we can also prevent it by lowering the line of the force $F$ closer to the floor, thereby reducing the tumbling couple).

The translational equation of motion is the same as in case A:

$$ma=F-F_f,$$

so we also obtain:

$$v=\bigg (\frac{F}{m}-\mu g \bigg )t.$$

The kinetic energy is now limited to translational energy:

$$K=\frac{mv^2}{2}.$$

But this is not equal to the total work done on the cube, as it doesn't take into account friction work.

The friction work is given by:

$$W_f=F_fx,$$

or:

$$W_f=\mu mgx.$$

Where $x$ is the displacement over the time $t$.

We know from above that the acceleration $a$ is:

$$a=\frac{F}{m}-\mu g,$$

and the displacement $x$ is:

$$x=\frac{at^2}{2}=\frac{1}{2}\bigg (\frac{F}{m}-\mu g \bigg )t^2=\frac{1}{2}vt.$$

So for $W_f$ we get:

$$W_f=\frac{1}{2}\mu mgvt.$$

The total work $W_B$ now done becomes:

$$W_B=\frac{mv^2}{2}+\frac{\mu mgvt}{2}.$$

We can now also compare $W_A$ and $W_B$ but it's rather a long derivation, so I'll only post the conclusion here. For:

$$F>\frac{7\mu mg}{2},$$

then:

$$W_A>W_B$$.

However, in this earlier answer I showed that for rolling without slipping the critical coefficient of friction $\mu_c$ is given by:

$$\mu_c=\frac{FI}{mg(I+mR^2)}$$

Extracting $F$ from that expression and with $I=\frac{2mv^2}{5}$ we then find that:

$$F=\frac{7\mu_c mg}{2},$$

which proves that:

$$\large{W_A=W_B}.$$

Note also that where $\mu>\mu_c$ then $\mu_c$ needs to be used in the equations of motion as otherwise we would be over-estimating the friction force and torque.

C) Conclusions:

All things being equal and in the presence of sufficient friction, the velocity (translation) of a sphere and a cube are the same. Both cases require the same amount of work but in the case of a (non-tumbling) cube part of that work is lost as friction work and thus not conserved.

Whenever one applies a sideways force trough the center of gravity of an object, that force has two components: 1) a direct force that tries to overcome friction and slide the object, and 2) a torque that uses friction to produce a rotation of the object by lifting its center of gravity over the leading edge.

A short, flat object will tend to slide because the torque needed to raise the center of gravity over the long edge is large compared to the force needed to overcome friction; the same object place on its short edge will tumble, because the torque needed to raise the center of gravity over that short edge is comparatively small.

Thus rolling friction must always be less than sliding friction, by definition, because for an object to roll the counter-force of friction that leads to torque must be smaller than the counter-force of friction that must be overcome for a slide.

A wheel, then, is simply a special case in which (ideally) the center of gravity of the object never needs to be lifted, and so the friction needed to induce a roll (ideally) is vanishingly small.

therefore it is easier to move sphere than cube of same weight.

This is a very common question asked by students like me in mechanical engineering. The friction is irrespective of area means the friction generated, the vector F (The letter with which we denote) will form whether area in contact is less or more

But the ability to stop is determined by the Number of friction vectors developed per area of contact. On the surface of a cube the magnitude of the friction vector will be the same as the one developed on the point of contact of the sphere(Irrespective of the surface of contact).

But the number of friction vectors produced per unit area will be different. This is one of the explanation given to me by our professor. Hence the cube requires more force than a sphere.

## protected by Qmechanic♦Nov 19 '15 at 16:33

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