Pulleys

Pulleys are well-known simple machines which can be used to reduce lifting force. The figure below shows (left to right) 1:1, 2:1, and 4:1 pulleys.

m FY

Xm

F

m

F

Let us derive the required lifting force for each pulley. The figure below shows (left to right) free-body diagrams (FBDs) for the 1:1, 2:1, and 4:1 pulleys.

F FW

R R

W

F1F1F1

F1F1

F1 F

R

W

F2 F2F2

1:1 Pulley. For the left FBD above, from static balance the cable tension lifting the weight must equal

the lifting force: mgWF . Then from equating the sum of Y forces to zero, mgWFR 2 . This pulley does not reduce the required lifting weight.

2:1 Pulley. For the left of the center two FBDs above, from static balance the cable tension F1 lifting

the weight is only half of the weight since one end of the rope is tied to the frame: mgWF 12 and so 2/1 mgF . The right of the center two FBDs above is similar to that of the 1:1 pulley; since the rope is

cut for drawing FBDs, the same F1 acts down on this FBD as shown (equal & opposite forces from Newtons Third Law). Therefore, the lifting force is 2/1 mgFF . Further, mgFFR 1 . The total reaction force on the frame is 2/31 mgFRRtot . This pulley cuts the required lifting weight in two.

4:1 Pulley. The left-most of the right three FBDs above is similar to the left FBD of the 2:1 pulley

above: mgWF 22 and so 2/2 mgF . Since the F2 rope is attached to the middle pulley, rope tension F1 may be found from: 212 FF and so 4/2/21 mgFF . The right-most of the right three FBDs above is similar to that of the 1:1 pulley; since the rope is cut for drawing FBDs, the same F1 acts down on this FBD as shown (equal & opposite forces from Newtons Third Law). Therefore, the lifting force is 4/1 mgFF . Further, 2/1 mgFFR . The total reaction force on the frame is

4/521 mgFFRRtot . This pulley cuts the required lifting weight in four.

Comparing the total reaction force on the frame for all three pulleys, we see that increased mechanical advantage also reduces the frame reaction force: this reaction force must be the weight, plus the lifting force F. As F is reduced, so is the total frame reaction.

Obviously, a simple machine cannot reduce work, it can only reduce the input force effort. The tradeoff for reduced F in these pulleys is more rope must be pulled: the 2:1 pulley must pull twice as much rope as the 1:1 pulley, and the 4:1 pulley must pull four times as much rope as the 1:1 pulley. In all three cases the work (force times distance) will theoretically be the same.

User chooses: 1:1, 2:1, or 4:1 pulley

Computer sets:

m = 10 kg, g = 9.81 m /s2, (in the Y direction).

Visualize: pulley lifting mass

Numerical Display:

F, R, F1, F2, Rtot

User Feels: Lifting force F

Examples: To lift the 10 kg mass 1 m against gravity (98.1 Nm of work), the results are:

1:1 Pulley 10.98F N, 20.196 RRtot N, 1ropeL m

2:1 Pulley 05.491 FF N, 10.98R N, 15.147totR N, 2ropeL m

4:1 Pulley 05.492 F N, 52.241 FF N, 05.49R N, 62.122totR N, 4ropeL m