Mr. Rogers' AP Physics C: Newton's Law Problems

Harness Type Elevator Problem
Bob still has a mass = 200 kg. He decides to install an elevator (of sorts in his house). The "elevator consists of a rope running over a pulley attached to the ceiling. To operate the make shift elevator, Bob ties the one of rope around his waist and pulls downward on the other dangling rope. This action makes him accelerate upward at a rate of 3 m/s2. What is the tension in the rope?

Solution

Define downward as negative. Here it's more convenient to draw a separate free body diagram. The dashed line box identifies the object in the free body diagram. Since 2 ropes protrude from the top of the dashed box, 2 tension forces are show on the free body diagram. Because ropes can only pull not push, the tension forces have to go upward. The pulley is modeled as massless and friction free, so the 2 tension forces have to be equal.

To solve the problem, we must find the normal force.

S F = ma

Ft1 + Ft2 + Fw  = ma

since Ft1 = Ft2

2 Ft + Fw = ma

2 Ft = ma - Fw

Ft = 1/2 (ma - Fw)

 but  Fw = - mg "g" is not an acceleration although it has the units of acceleration. It is an indication of the gravity field. "g" is a vector and has a negative value since it points downward. On planet Earth g = 9.8 m/s2. An object will have this downward acceleration due to gravity only when it is in freefall.

Ft = 1/2 [ ma - (- mg) ]

 = 1/2 [ ma + mg ] The mg term will always be positive but the ma term's sign will change depending on whether the acceleration points up or down.

= 1/2 m(a + g)

 = 1/2 (100 kg)( 3 m/s2 + 9.8 m/s2 ) a = 3 m/s2 because the acceleration is upward. It would be negative if the acceleration were downward

= 640 n

Conclusion and Significance

Using the pulley and harness reduces the required tension force by 1/2. giving Bob a mechanical advantage of 2. The system, however, has a downside if Bob pulls the rope downward by 1 meter, he will go upward by only 1/2 a meter. In other words, there is no "free lunch". Bob will have to pay for the increase in total upward force by going upward half as fast.

Mr

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