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# pulley

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Name_______________________
Investigation: Mechanical Advantage of a pulley system
Theory: A pulley system can be used to reduce the force needed to get a job done (lifting a
weight) at the expense of increasing the distance through which the force must act. If an object
is lifted directly, a force equal to the weight of the object must be applied. If a pulley system
with one or more movable pulleys is used, the force needed is (much) less than the weight of the
object. The mechanical advantage (M.A.) is the ratio of the weight of the object lifted to the
pulling force needed. The ideal mechanical advantage (I.M.A.) is the ratio of the distance
through which the pulling force is exerted to the distance through which the object moves.
F                                d
M . A. out                    I . M . A. in
Fin                              d out
Objective: Determine the actual mechanical advantage and the ideal mechanical advantage of
two different pulley systems.

Materials: pulleys, string, masses, spring scales, ring stand, clamp (support)

Procedure:
1. Adjust spring scale to zero! 2. I said, Adjust spring scale to zero! 3. Are you sure the scale is
adjusted to zero? 4. Set up pulley system #1. 5. Lift the weight by pulling vertically upward
with the spring scale at a constant speed. Record the spring scale reading as Fin. Weigh the
object with the spring scale and record as Fout. 6. Now move the spring scale through a
measured vertical distance and note the corresponding distance that the object moves. Record
these two distances as din and dout respectively. Choose as large a distance as practical for din to
reduce the percent uncertainty. Note that the I.M.A. depends only on these distances and can be
predicted visually by counting supporting strings in the pulley system. If the I.M.A. from your
distances does not match your predicted value, get help! 7. Set up pulley system #2 8,9.
Repeat steps 5 and 6 for pulley system #2.

Data:
mass being lifted by pulley system _________g = __________kg

weight of mass being lifted by pulley system ___________N

system #1      Fin(N)____________ Fout(N)____________           M.A.______________

din(cm)___________ dout(cm) ___________ I.M.A._____________

system #2      Fin(N)____________ Fout(N)____________           M.A.______________

din(cm)___________ dout(cm) ___________ I.M.A._____________
Calculations: The "efficiency" of a machine is a measure of how much work you get out,
compared to the energy you put in. The "useful work" in this case is lifting the mass (force x
distance) where the force is the weight of the mass. The Energy you put in is the work you do
lifting the end of the string where the scale is.
Energy  W  Fd

where the force is the reading on the scale in Newtons. A "perfect" machine would give the
same amount of work out as energy in. Real machines use moving parts that all have friction (it
takes energy to overcome frictional forces), and all have mass (it takes energy to move them), so
the output work is less than the input work. We measure efficiency as this ratio:

useful _ Work _ out
E
Energy _ in

So the efficiency will always be a number between 0 and 1 (you can multiply by 100 to get a
percent). From your data, complete the table below. Remember to convert your cm distances to
meters so you get Nm (Joules) for the energy units.

system #1      Useful Work out(J)___________              Energy in(J)___________

Efficiency____________                  M.A./I.M.A______________

system #2      Useful Work out(J)___________              Energy in(J)___________

Efficiency____________                  M.A./I.M.A______________

Questions:
1) What two factors might cause the M.A. to be less than the I.M.A.?

2) Who killed Cock Robin?
3) Can the M.A. ever be greater than the I.M.A.? Explain.

4) What are the units for Efficiency?
5) Which system was more efficient and why?

System #1                                             System #2

support
spring
scale

support

spring
scale

weight                                         weight

Extra credit: Show (with equations) that the ratio M.A to I.M.A is the same as the efficiency.

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