Question

If the coefficient of static friction between a car’s tires and the pavement is 0.65, calculate the minimum torque that must be applied to the 66-cm-diameter tire of a 1080-kg automobile in order to “lay rubber” (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equal share of the weight.

Short answer

The minimum torque that will make the tire slip is \(567.6\;{\rm{m}}\;{\rm{N}}\).

Step-by-step solution

Identification of the given data

The mass of the car is \(m = 1080\;{\rm{kg}}\).

The coefficient of static friction is \({\mu _{\rm{s}}} = 0.65\).

The diameter of the tire is \(d = 66\;{\rm{cm}} = 0.66\;{\rm{m}}\).

Definition of static friction

The friction experienced when a person tries to move a stationary object on a surface without actually triggering any relative motion between the body and the surface on which it is moving is known as static friction.

Definition of torque

Torque is defined as the measure of the force that causes an object to rotate about an axis. It is also called the rotational equivalent of linear force.

Mathematically, torque is represented as:

\(\begin{aligned}{c}\vec \tau = \vec r \times \vec F\\\left| \tau \right| = rF\sin \theta \end{aligned}\) … (i)

Here, F is the applied force, and r is the perpendicular distance between the point about which the torque is calculated and the point of application of force.

Calculation of the force of static friction

The tires of the car will not slip as long as the force F between the tires and the road is less than the maximum static friction. You need to equate this minimum force F to the maximum value of static friction.

\(F = {\mu _{\rm{s}}}N\).

Here, N is the normal force that the pavement applies on the tire to keep the car at rest. It is equal to the weight (force of gravity) acting downward. The mass of each tire is \(\frac{1}{4}th\) of the total mass of the car. Therefore,

\(\begin{aligned}{c}F = {\mu _{\rm{s}}}\left( {\frac{m}{4}} \right)g\\ = \left( {0.65} \right)\left( {\frac{{1080\;{\rm{kg}}}}{4}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\\ = 1720\;{\rm{N}}\end{aligned}\)

Calculation of the torque that must be applied on the tire

The torque that must be applied to make the tire slip is given by:

\(\begin{aligned}{c}\tau = rF\\ = \left( {\frac{{0.66\;{\rm{m}}}}{2}} \right)\left( {1720\;{\rm{N}}} \right)\\ = 567.6\;{\rm{m}}\;{\rm{N}}\end{aligned}\)