Question

The pulley system shown in the figure is used to lift a 52-{rm kg} crate. Note that one chain connects the upper pulley to the ceiling and a second chain connects thelower pulley to the crate. Assume the masses of the chains, pulleys, and ropes are negligible.
Determine the force F required to lift the crate with constant speed.

Answer 1

to find force
Fnet=F+F-mg
F=Mg/2

52(9.80)/2 =254.8 N, which rounds down to 250 N.

Answer 2

The pulley system shown in Figure 6-25 is used to lift a 39 kg crate. Note that one chain connects the upper pulley to the ceiling and a second chainconnects the lower pulley to the crate. Assume that the masses of the chains, pulleys, and ropes are negligible and the crate is rising with anacceleration of 2.9 m/s2.

Weight of crate = 39 * 9.8 = 382.2 N
The second chain connects the lower pulley to the crate and the crate is rising with an acceleration of 2.9 m/s2.

The tension in the lower chain is pulling the crate up.
The weight is pulling the crate down.
The sum of these 2 forces are causing the crate to accelerate up at 2.9 m/s^2

Tension lower – Weight = mass * acceleration
Tension lower – 382.2 = 39 * 2.9
Tension lower = 382.2 + (39 * 2.9)
Tension lower = 495.3 N

I can not see Figure 6-25, so, I assume the rope is pulled down to lift the crate up. This means the force of the rope is ½ of the tension in the lowerchain.
Force of rope = ½ * 495.3 = 247.65 N

The upper chain connects the upper pulley to the ceiling.
So, the upper chain supports both pulleys, the rope, and the crate.

Two forces are pulling down, the force of the rope and the weight of the crate.
The tension in the upper chain is pulling up.

The tension in the upper chain = Force of rope + Weight of crate
The tension in the upper chain = 247.65 + 382.2 = 629.85 N



(a) Determine the force. 247.65 N