If the coefficient of static friction between tires and pavement is 0.7, calculate the minimum torque that must be applied to the 63-cm-diameter tire of a 925-kg automobile in order to "lay rubber" (make the sheels spin, slipping as the car accelerates). Assume each wheel [is a disk that] supports an equal share of the weight.

weight on each tire = 945*9.8/4 = 2315 Newtons on each tire

force of friction at slip = .7*2315 = 1621 N
times radius (0.63/2) to get torque

Why did the tire go to therapy?

Because it was feeling a little "unbalanced"!

Alright, let's get down to business. To calculate the minimum torque required to make the wheels of the car slip, we need to consider the maximum friction force that can be generated between the tires and pavement.

The maximum friction force (F_friction) can be calculated using the equation F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Given that the car's weight is evenly distributed among all four tires, each tire supports 925 kg / 4 = 231.25 kg. The normal force acting on each tire can be calculated as N = m * g, where m is the mass of the car and g is the acceleration due to gravity.

Now, let's plug in the values:
N = 231.25 kg * 9.8 m/s² = 2264.75 N

Since there are four tires, each tire will have a maximum friction force of F_friction = 0.7 * 2264.75 N = 1585.32 N.

To calculate the torque required, we'll use the equation T = F * r, where T is the torque, F is the force, and r is the radius of the tire. The radius of the tire in meters is 63 cm / 100 = 0.63 m.

Now, let's calculate the torque:
T = F_friction * r = 1585.32 N * 0.63 m = 997.01 N⋅m.

So, to "lay rubber" and make the tires of the car slip, a minimum torque of approximately 997.01 N⋅m would need to be applied. Let's hope the car enjoys this spinning performance!

To calculate the minimum torque required to make the tires "lay rubber" and slip, we'll need to take into account the frictional force between the tires and the pavement.

First, let's calculate the weight supported by each tire. Since each wheel supports an equal share of the weight, we can divide the total weight of the automobile by the number of wheels:

Weight supported by each tire = Total weight of the automobile / Number of wheels
Weight supported by each tire = 925 kg / 4 (assuming a standard 4-wheel car)
Weight supported by each tire = 231.25 kg

Next, we need to calculate the force of friction between the tires and the pavement when they are on the verge of slipping. This can be done using the equation:

Frictional force = Coefficient of static friction * Normal force

Given that the coefficient of static friction is 0.7 and the weight supported by each tire is 231.25 kg, we can substitute these values into the formula:

Frictional force = 0.7 * 231.25 kg
Frictional force = 161.88 N

Now, let's calculate the torque required to produce this force of friction. The torque can be calculated using the equation:

Torque = Force * Radius

The radius of the tire can be obtained by dividing its diameter by 2:

Radius = Diameter / 2
Radius = 63 cm / 2
Radius = 31.5 cm = 0.315 m

Plugging in the values for the frictional force and the radius into the torque equation, we get:

Torque = 161.88 N * 0.315 m
Torque = 50.96 N*m

Therefore, the minimum torque that must be applied to the 63-cm-diameter tire of a 925-kg automobile in order to "lay rubber" is approximately 50.96 N*m.

To calculate the minimum torque required to make the tires spin (slip) on a 925-kg automobile with a 63-cm-diameter tire, we need to consider the forces involved.

The maximum static friction force can be calculated using the formula:

Fs_max = μs * N

where μs is the coefficient of static friction and N is the normal force.

First, let's calculate the normal force acting on each tire. Since each wheel supports an equal share of the weight, the weight is divided equally between the four wheels. So, the normal force can be calculated as:

N = (weight of the automobile) / (number of tires)

N = (925 kg * 9.8 m/s^2) / 4

N = 2258.25 N

Now, we can calculate the maximum static friction force for each tire:

Fs_max = 0.7 * 2258.25 N

Fs_max = 1580.775 N

To make the wheels slip, the torque applied must overcome the static friction force. The torque required can be calculated using the formula:

Torque = (frictional force) * (radius of the tire)

Torque = (Fs_max) * (0.5 * diameter of the tire)

Note that the radius of the tire is half of the diameter.

Torque = (1580.775 N) * (0.5 * 63 cm)

Torque = (1580.775 N) * (0.5 * 0.63 m)

Torque = 498.462 N*m

Therefore, the minimum torque that must be applied to the 63-cm-diameter tire of the 925-kg automobile in order to "lay rubber" is approximately 498.462 N*m.