Figure 1 ) The car shown in the figure has mass m (this includes the mass of the

Question

Figure 1) The car shown in the figure has mass m (this includes the mass of the wheels). The wheels have radiusr, mass m_w, and moment of inertiaI=km_w r^2. Assume that the axles apply the same torquet to all four wheels. For simplicity, also assume that the weight is distributed uniformly so that all the wheels experience the same normal reaction from the ground, and so the same frictional force.

a. If there is no slipping, a frictional force must exist between the wheels and the ground. In what direction does the frictional force act? Take the positive x direction to be to the right.

b.positive x directio

a. negative x direction

b.positive x directio

part b. Use Newton's laws to find an expression for the net external force acting on the car. Ignore air resistance. Express your answer in terms of any given variables and f, the force of friction acting on each wheel. part c: Use Newton's laws to find an expression for N, the normal force on each wheel. Express your answer in terms ofm and g, the magnitude of the acceleration due to gravity. part d: Now assume that the frictional forcef is not at its maximum value. What is the relation between the torquet applied to each wheel by the axles and the accelerationa of the car? Once you have the exact expression for the acceleration, make the approximation that the wheels are much lighter than the car as a whole. Express your answer in terms of some or all of the variables m, r,t and the magnitude of the acceleration due to gravity g.

Explanation / Answer

A. The frictional force is in the positive X direction
B. The net force is in the horizontal direction, f is the friction force in each wheel. so the answer is Fnet=4f
C. The weight is distributed equally on each wheel in the Y direction, gives us the formula 4Nwheel-mg=0 with a final answer of N=mg/4
D. torque force at each wheel is tau=rf, linear acceleration is obtained from the sum of the forces 4f=ma, f=tau/r and f=ma/4, (ma/4=tau/r) giving us the final answer of a=4tau/mr    (please note that the tau symbol is in the drop down list and looks like a cross between a lowercase r and an upper case T)

I just did this problem and the answers were confirmed correct.
A. The frictional force is in the positive X direction
B. The net force is in the horizontal direction, f is the friction force in each wheel. so the answer is Fnet=4f
C. The weight is distributed equally on each wheel in the Y direction, gives us the formula 4Nwheel-mg=0 with a final answer of N=mg/4
D. torque force at each wheel is tau=rf, linear acceleration is obtained from the sum of the forces 4f=ma, f=tau/r and f=ma/4, (ma/4=tau/r) giving us the final answer of a=4tau/mr    (please note that the tau symbol is in the drop down list and looks like a cross between a lowercase r and an upper case T)

I just did this problem and the answers were confirmed correct.
A. The frictional force is in the positive X direction
B. The net force is in the horizontal direction, f is the friction force in each wheel. so the answer is Fnet=4f
C. The weight is distributed equally on each wheel in the Y direction, gives us the formula 4Nwheel-mg=0 with a final answer of N=mg/4
D. torque force at each wheel is tau=rf, linear acceleration is obtained from the sum of the forces 4f=ma, f=tau/r and f=ma/4, (ma/4=tau/r) giving us the final answer of a=4tau/mr    (please note that the tau symbol is in the drop down list and looks like a cross between a lowercase r and an upper case T)

I just did this problem and the answers were confirmed correct.

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