The system in Figure 1 consists of a pulley that fits loosely over a shaft. The shaft is rigidly supported at its ends and does not turn. The diameter of the wheel is 150 mm, and that of the shaft is 20 mm. between the shaft and the pulley the coefficient of static friction is 0.40, and the

Question

ANALYSIS OF A PULLEY SYSTEM WITH BEARING FRICTION

The system in Figure 1 consists of a pulley that fits loosely over a shaft. The shaft is rigidly supported at its ends and does not turn. The diameter of the wheel is 150 mm, and that of the shaft is 20 mm. between the shaft and the pulley the coefficient of static friction is 0.40, and the coefficient of kinetic friction is 0.35.
(a) Determine the minimum tension in the rope to hold the 300-N block in a stationary position.
(b) Determine the minimum tension in the rope to just start raising the block. Also express this answer as a percentage of the tension found in (a).

Accepted Answer

Goal Find the tension T in the rope to keep the weight stationary (a), and to just begin raising it (b).
Given We are given the diameters of the pulley (150 mm) and the shaft (20 mm) on which it is loosely fit. In addition, we are given the coefficients of static ([latex] \mu _{s} [/latex] = 0.40) and kinetic ([latex] \mu _{k} [/latex] = 0.35) friction and the weight of block (300 N) being held by the pulley system.
Assume The weight of the pulley is negligible and that the rope does not slide in the pulley groove.
Draw For (a) First we draw the free-body diagram of the pulley in a stationary position (Figure 2a). The clockwise moments created about a moment center at A by the rope tension and the bearing friction are balanced by the counterclockwise moment created by the block’s weight about A. bearing friction in the clockwise direction means that it is helping to keep the block stationary (and therefore, the required tension T is at its minimum).
Formulate Equations and Solve For (a) From (7.18), the moment created by the bearing friction is M = r • μF, where F is the magnitude of the force supported by the bearing and r is the radius of the shaft. For the situation in Figure 2a, the force supported by the bearing is [latex] A _{y} [/latex] , and it is appropriate for [latex] \mu _{s} [/latex] to use the static coefficient of friction μs (because the pulley is not turning relative to the shaft for part (a)). Therefore,

[latex]M_{bearning friction}= r_{shaft} \bullet \mu _{s} A_{y}[/latex] (1)

Now we find the value of Ay by using equilibrium:

[latex]\sum{F_{y}\left(\uparrow + \right) } =A_{y}- 300 N - T=0[/latex]
[latex] A_{y}=300 N + T[/latex] (2)

Substituting (2) into (1), we find that the magnitude of the moment created by the bearing friction is
[latex]M_{bearning friction}= r_{shaft} \bullet \mu _{s} \left(300 N + T\right)[/latex]
based on equilibrium and the free-body diagram in Figure 2 we write:

[latex]\sum{M_{z @ A} } \left(\curvearrowleft + \right)=300 N \left\lgroup\frac{150}{2} mm \right\rgroup - T\left\lgroup\frac{150}{2} mm\right\rgroup - \underbrace{ M_{bearning friction}}_{from(3)} =0[/latex] (4)
[latex]300 N \left\lgroup\frac{150}{2} mm \right\rgroup - T\left\lgroup\frac{150}{2} mm\right\rgroup - r_{shaft}\bullet \mu _{s} \left(300 N + T\right)=0[/latex]

Rearranging and substituting in for r = 10 mm and [latex] \mu _{s} [/latex] = 0.40, we solve for the minimum tension T in the rope to hold the 300-N block in a stationary position:

T = 270 N

Draw For (b) we draw the free-body diagram of the pulley just as the block is about to start moving upward (Figure 2b). The clockwise moment created about a moment center at A by the rope tension is balanced by counterclockwise moments created by the block’s weight and the bearing friction. The moment created by the bearing friction being in the counterclockwise direction means that it is working to prevent the impending upward movement of the block.
Formulate Equations and Solve For (b), as in the calculations for (a), the magnitude of the moment created by the bearing friction is

[latex]M_{bearning friction}= r_{shaft} \bullet \mu _{s} \left(300 N + T\right)[/latex]

but its direction is in the counterclockwise direction. It is appropriate to use the static coefficient of friction because the block is just about to start moving upward.
based on equilibrium and Figure 2b we write:

[latex]\sum{M_{z @ A} } \left(\curvearrowleft + \right)=300 N \left\lgroup\frac{150}{2} mm \right\rgroup - T\left\lgroup\frac{150}{2} mm\right\rgroup + M_{bearning friction}=0[/latex] (5)
[latex]300 N \left\lgroup\frac{150}{2} mm \right\rgroup - T\left\lgroup\frac{150}{2} mm\right\rgroup + r_{shaft}\bullet \mu _{s} \left(300 N + T\right)=0[/latex] (6)

Rearranging and substituting in for [latex]r_{shaft}[/latex] and [latex] \mu _{s} [/latex] , we solve for the minimum tension T in the rope when the block is just about to start moving upward.

T = 334 N

This is 23.8% greater than the tension required to hold the block stationary.
Check The results are summarized in Table 1.

Table 1 Summary of Results

Condition	Tension
block stationary	270 N
block just starting to move upward	334 N

These results say that
• more tension is required to just start raising the block. This makes sense, since the moment created by the tension must balance the moments created by the block’s weight and the bearing friction.
• less tension is required to hold the block stationary. This makes sense, since the moments created by tension and bearing friction balance the moment created by the block’s weight.

Question 7.2.1: The system in Figure 1 consists of a pulley that fits loosel...

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