Treadmill’s Belt Drive An electric motor is used to power an exercise treadmill. Forces are applied to the treadmill’s shaft by the motor’s drive belt and by the wide, fl at belt that is the surface used for walking or running. The tight and loose spans of the drive belt together apply 110 lb to

Question

Treadmill’s Belt Drive

An electric motor is used to power an exercise treadmill. Forces are applied to the treadmill’s shaft by the motor’s drive belt and by the wide, flat belt that is the surface used for walking or running. The tight and loose spans of the drive belt together apply 110 lb to the shaft, and the treadmill’s belt applies 70 lb. The shaft is supported by ball bearings on each side of the belt. Calculate the magnitudes and directions of the forces exerted by the shaft on the two bearings. (See Figure 4.28 on page 162)

Approach

We are tasked with finding the forces on the two bearings from the two belts. We first assume that all the forces act parallel to the y-direction. The free body diagram of the shaft is drawn, along with the sign conventions for the coordinate directions and rotation. On the diagram, we first label the 110-lb and 70-lb belt tensions, and then we denote the forces exerted by the bearings on the shaft as [latex]F_A [/latex] and [latex]F_B[/latex]. At this point, we don’t know whether those unknown forces act in the positive or negative y-directions. By drawing them on the free body diagram using our sign convention, we will rely on the calculation to determine the actual direction of the forces. If a numerical value turns out to be negative, the result will mean that the force acts in the negative y-direction. (See Figure 4.29)

Accepted Answer

Because there are two unknowns ( [latex]F_A [/latex] and [latex]F_B[/latex] ), two equilibrium equations are needed to solve the problem. By summing forces in the y - direction

[latex](110 lb) - (70 lb) + F_A + F_B = 0 \longleftarrow [\sum\limits_{i=1}^{N}{F_{y,i}=0} ] [/latex]

or [latex]F_A [/latex] + [latex]F_B[/latex] = - 40 lb. We will apply a moment balance for the second equation. By choosing the pivot point to coincide with the center of bearing A, force [latex]F_A[/latex] will be eliminated from the calculation. Summing moments about point A, we have

[latex](110 lb)(4 in.)- (70 lb)(19 in.) + (F_B)(36 in.) = 0 \longleftarrow [\sum\limits_{i=1}^{N}{M_{A,i}=0} ] [/latex]

and [latex]F_B[/latex] = 24.72 lb. The motor belt’s tension and[latex] F_B [/latex] exert counterclockwise (positive) moments about A, and the treadmill belt’s tension balances those components with a negative moment. Substituting this value for [latex]F_B[/latex] into the force balance returns [latex]F_A[/latex] = - 64.72 lb.

Discussion
These forces are the same order of magnitude as the applied forces, and that makes sense. Also, when in use, the treadmill will exert forces on the bearings in the x- and[latex] z[/latex]-directions, but those are not accounted for in our analysis. Since the calculated value for [latex]F_A[/latex] is negative, the force that is exerted by bearing A on the shaft acts in the negative y-direction with magnitude of 64.72 lb. Following the principle of action–reaction from Newton’s third law, the directions of the forces exerted by the shaft on the bearings are opposite those exerted by the bearings on the shaft.

Bearing A: 64.72 lb in the negative y-direction

Bearing B: 24.72 lb in the positive y-direction

Question 4.8: Treadmill’s Belt Drive An electric motor is used to power an...

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