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[pizza1994]

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A laborer expends 800 J to lift a block to a height h. He then repeats the task using a simple non-motorized pulley system that reduces by half the input force he must provide. With the pulley system in operation, how much work must the laborer perform in order to lift the block to height h?

Source: TPR
asnwer: 800 J

why is it 800 J?

 

[t5Nitro]

A laborer expends 800 J to lift a block to a height h. He then repeats the task using a simple non-motorized pulley system that reduces by half the input force he must provide. With the pulley system in operation, how much work must the laborer perform in order to lift the block to height h?

Source: TPR
asnwer: 800 J

why is it 800 J?

Work in = Work out.

Fd(in) = Fd(out)

What happens when the force is reduced by 1/2 due to the pulley he is operating? The same overall work is used to lift the block, therefore the same amount of work must be input to raise that block. With 1/2 the force, he pulls the pulley 2x as far to balance the equation.

Example:

Before using the pulley:

Takes the worker 800 N to move the block 1 meter (Work = Fd) so work = (800 N)(1 m) = 800J

After using the pulley:

Reduce the force required by 1/2 --> (400 N)(2 m) = 800 J

In this last case, he just had to move the block twice as far with less force.

Overall, the block was raised to the same height and by PE = mgh, you know the same amount of joules was performed in both cases.

 

[pizza1994]

Work in = Work out.

Fd(in) = Fd(out)

What happens when the force is reduced by 1/2 due to the pulley he is operating? The same overall work is used to lift the block, therefore the same amount of work must be input to raise that block. With 1/2 the force, he pulls the pulley 2x as far to balance the equation.

Example:

Before using the pulley:

Takes the worker 800 N to move the block 1 meter (Work = Fd) so work = (800 N)(1 m) = 800J

After using the pulley:

Reduce the force required by 1/2 --> (400 N)(2 m) = 800 J

In this last case, he just had to move the block twice as far with less force.

Overall, the block was raised to the same height and by PE = mgh, you know the same amount of joules was performed in both cases.

how is the block raised to the same height? before pulley the height was 1 m and then after pulley the height was 2 m. I dont see why the asnwer has to still be 800 J if i want to lift it to 1 m!!!

 

[gettheleadout]

In this last case, he just had to move the block twice as far with less force.

Overall, the block was raised to the same height and by PE = mgh, you know the same amount of joules was performed in both cases.

No, the block moves the same distance. The lower force had to be applied over a greater distance.

how is the block raised to the same height? before pulley the height was 1 m and then after pulley the height was 2 m. I dont see why the asnwer has to still be 800 J if i want to lift it to 1 m!!!

The value d for W = Fd is not the distance the block moves, it's the distance over which the force is applied. If you have a pulley system that cuts the required input force in half, you'll have to pull twice the length of rope down to lift a mass to the same height as without the pulley's mechanical advantage. The block moves a distance h in both cases.

 

[pizza1994]

No, the block moves the same distance. The lower force had to be applied over a greater distance.

The value d for W = Fd is not the distance the block moves, it's the distance over which the force is applied. If you have a pulley system that cuts the required input force in half, you'll have to pull twice the length of rope down to lift a mass to the same height as without the pulley's mechanical advantage. The block moves a distance h in both cases.

yeah so that makes sense that if you have a pulley that requires half the input force then you will have to pull twice the lengh of rope....but then I dont get how using a pulley is useful if the amount of work done is always the same with or without a pulley. In my mind a pulley is useful if it is able to reduce the amount of work/energy required to lift something a certain height

 

[gettheleadout]

yeah so that makes sense that if you have a pulley that requires half the input force then you will have to pull twice the lengh of rope....but then I dont get how using a pulley is useful if the amount of work done is always the same with or without a pulley. In my mind a pulley is useful if it is able to reduce the amount of work/energy required to lift something a certain height

Think about it like this. Imagine I can only deadlift 200 pounds, but I need to get a 400 pound crate up six inches up off the floor so someone else can slide a rolling cart under it. Now, if I try to lift it straight up I'm obviously not strong enough. So let's say I go get a rope and loop it over a pulley overhead. If the rope only runs up and over a pulley, the mechanical advantage is only 1; the force I apply pulling down will be directly transferred as upward force to the crate. Since I don't weigh 400 pounds, I can't exert enough downward force to lift the crate up using the rope. Now let's say I loop the rope over more pulleys and hooks on the crate and get a mechanical advantage of 4. Now I only have to exert 100 pounds of force (as tension on the rope) to get 400 pounds of total force output by the multiple segments of rope pulling on the crate. Since I weigh more than 100 pounds, I can easily pull on the rope to lift the 400 pound crate. However, because of conservation of energy and the work-energy theorem, I can only put as much energy into the crate (as gravitational potential energy) as I can exert in lifting it. This means that by multiplying the force the crate feels I sacrifice the 1:1 relationship of my movement distance : the crate's movement distance. Now, to lift the crate six inches up, I have to pull the rope 24 inches down.

 

[pizza1994]

Think about it like this. Imagine I can only deadlift 200 pounds, but I need to get a 400 pound crate up six inches up off the floor so someone else can slide a rolling cart under it. Now, if I try to lift it straight up I'm obviously not strong enough. So let's say I go get a rope and loop it over a pulley overhead. If the rope only runs up and over a pulley, the mechanical advantage is only 1; the force I apply pulling down will be directly transferred as upward force to the crate. Since I don't weigh 400 pounds, I can't exert enough downward force to lift the crate up using the rope. Now let's say I loop the rope over more pulleys and hooks on the crate and get a mechanical advantage of 4. Now I only have to exert 100 pounds of force (as tension on the rope) to get 400 pounds of total force output by the multiple segments of rope pulling on the crate. Since I weigh more than 100 pounds, I can easily pull on the rope to lift the 400 pound crate. However, because of conservation of energy and the work-energy theorem, I can only put as much energy into the crate (as gravitational potential energy) as I can exert in lifting it. This means that by multiplying the force the crate feels I sacrifice the 1:1 relationship of my movement distance : the crate's movement distance. Now, to lift the crate six inches up, I have to pull the rope 24 inches down.

ahh right the whole deadlift concept worked well. Basically when I cant deadlift more than a certain amount because Im not strong enough then thats when you use pulleys....there used to lift extremely heavy thing at the cost of pulling down on the rope several inches...but in the end you would rather pull down 24 inches instead of trying to lift something wayyy heavier than you!!!!

THANKS 🙂