# What happens to acceleration upon exceeding static friction?

1. Jul 4, 2015

### Ocata

So if I compare how the accelerations are observed in two objects, a car and a block, both having the same mass of 100kg on a surface with $μ_{s} = .5$ and $μ_{k} = .1$

Car:

The car will have Static Friction Force $f_{s} = μ_{s}F_{N}=[μmg = ma] = [.5(100kg)(10m/s^{2}) = 100kg(a)]$ =>
so acceleration will be $==> [itex] \frac{f_{s}}{m}=\frac{μ_{s}F_{N}}{m}= \frac{μmg}{m} = \frac{.5(100kg(10)}{100kg}= \frac{500N}{100kg} = 5m/s^{s} = a = acceleration$

So the force on the car would need to be greater than 500N. And the acceleration would need to be 5m/s^2.

Block:

The block will have Static Friction Force $f_{s} = μ_{s}F_{N}=[μmg = ma] = [.5(100kg)(10m/s^{2}) = 100kg(a)]$ =>
so acceleration will be $==> \frac{f_{s}}{m}=\frac{μ_{s}F_{N}}{m}= \frac{μmg}{m} = \frac{.5(100kg(10)}{100kg}= \frac{500N}{100kg} = 5m/s^{s} = a = acceleration$

So the force on the block would also need to be greater than 500N. And the acceleration would need to be 5m/s^2.

So, no difference of between the car and block in terms of how much Force and acceleration need to be applied in order to break static friction.

However, there seems to be a visual difference that is confusing.

It looks as though as Force is being applied to the car and the block from 0 to 500N, there is an associated acceleration applied to the car and block from 0 to 2m/s^2. The difference is, the car can be viewed as literally accelerating while the block remains stationary. So the "a" in the formula [μmg = ma] seems to be manifesting into the vehicles accelerating motion. However, although the same force is being applied to the block, you don't see the force having any affect on the blocks motion.

So, my question now is, what happens to the acceleration of an object upon breaking past the maximum static friction?

Would I measure the block accelerating at a rate of just over 2m/s^2 exactly at the instant it breaks from static friction? Would I measure the acceleration quickly increase from 0 to 2m/s^2? Will it immediately drop to the net acceleration of $\frac{f_{s}-f_{k}}{m} = a_{net}$? Or will the acceleration at the instant of breaking free from the static friction automatically start at $\frac{f_{s}-f_{k}}{m} = a_{net}$?

Last edited: Jul 4, 2015
2. Jul 4, 2015

### Staff: Mentor

First and foremost, you're not using 'acceleration' correctly. You don't apply an acceleration. You apply a force and if that force results in a change in the velocity of the object, then the object is accelerating.

Static friction applies a force that resists other applied forces. So if you push on a block, the force of static friction pushes in the opposite direction. The force that static friction applies is equal to the force you apply until the force you apply exceeds a value equal to μsmg. At that point the block begins to accelerate and kinetic friction applies instead of static friction. Kinetic friction applies a force on the block that points in a direction opposite to the blocks motion and the magnitude of the force is given by μkmg, where μk is less than or equal to μs (almost always less than).

The acceleration of the block is found by summing together all the forces acting on the block. If the sum is not zero then there is a net force being applied and the block will accelerate. In our example the sum of forces on the block is equal to zero until the force you are applying exceeds what static friction can apply.

The acceleration goes from 0 to whatever value it needs to be given the difference between the pushing force and the force of kinetic friction.

Remember that the wheels can turn, so the situation isn't the same as pushing a block.

That formula is not correct because you need to have the net force on the left, not just the force from friction. Either Fnet = MA or Fp + μmg = MA. Fp and μmg point in opposite directions so you'll actually have to subtract them to find the net force, which then determines the acceleration.

Last edited: Jul 8, 2015
3. Jul 8, 2015

### Ocata

Thank you Sir.