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

Elevator Problem: Bob has a mass = 200 kg. He has been told that he can lose weight by descending in an elevator. He places a bathroom scale in the elevator, stands on it, and presses the down button causing him to  descend at an acceleration of 4 m/s2. What does the bathroom scale read on the way down?

Solution

Define downward as negative. For convenience we draw the forces on the diagram and it becomes the free body diagram. Note that the bathroom scale will indicate the normal force acting upward on Bob. Since the acceleration is downward we know that the normal force has to be less than the weight force so that a net downward force acts on Bob.

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

S F = ma

Fn + Fw = ma

Fn = 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.

Fn = ma - (- mg)

 
     = 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.

      = m(a + g)

 
      = (100 kg)( - 4 m/s2 + 9.8 m/s2 ) a = - 4 m/s2 because the acceleration is downward. It would be positive if the acceleration were upward.

      = 580 n
 
Conclusion and Significance

Bob's original scale reading, when not accelerating, would have been mg or 980 n. While accelerating downward, Bob will think he's lost weight but in reality his mass and therefore his downward weight force will not have changed.

Will Bob feel lighter while accelerating downward? Yes! The sensation of weight depends on the normal force, not the weight force.

 

Metacognition Question

When accelerating downward, the scale reading will always be less than the scale reading of mg at rest. When accelerating upward the scale reading will always be more than the scale reading at rest.

If you get the opposite in your answer, what's wrong? Answer: you've put the wrong sign on the acceleration when you substituted it into your equation.