# Homework Help: Mechanical advantage machines

1. Jan 22, 2014

### lovelyrwwr

I've been at this problem for a good hour now and I just can't figure it out

Please help me think through this! It is a very difficult problem in my opinion. Or maybe I'm just missing a fundamental point...Thanks in advance!

-----------------------------------------------------------------------------------------

An engineer with a mass of 100 kg has designed an ideal mechanical advantage machine below. Platforms 1 and 2 are attached to the machine. When he steps on platform 1, platform 2 rises straight up. The maximum weight that he can lift using his machine in this manner is twice his own. The mechanical advantage of the machine cannot be adjusted. Platform 1 can be lowered a maximum of 10 m.

View attachment untitled.bmp

If the mass m is 100 kg, what are the forces upward on the engineer and the mass m, respectively?

The correct answer is: 600 N, 1200 N.

------------------------------------------------------------------------------------------
My thoughts:

So far I understand that because this is a machine, work will be the same, but the force you apply is smaller because you apply it over a greater distance. W = Fd.

Thus, the engineer must "pay the price" for lifting double his weight by applying the force of his weight over twice the distance. Thus, if the force (his weight, 1000 N) is applied for the 10 meters, then he can lift twice his weight (2000 N) to a maximum height of 5 meters.

--------------------------------------------------------------------------------------------

I also know that the engineer's acceleration is twice as large as the acceleration of the mass.

So my equations for the engineer is as follows:

(MASSman)(gravity) - FN = (MASSMAN)(2a)

And for the mass:

FN - (MASSbox)(gravity) = (MASSbox)a

This is obviously wrong. Some yahoo forum discussing this problem suggested that the normal force the box feels is twice the normal force that the engineer experiences...Is that correct? If so, I don't intuitively understand why that is the case.

2. Jan 23, 2014

### tiny-tim

hi lovelyrwwr!

i'm not sure you're understanding the question correctly

if m is 200 kg, then the situation is balanced: the 1000 N weight of the man balances the 2000 N weight of m so the acceleration is zero, and so the upward forces would be equal to the weights

if m is only 100 kg, the weights are as before,
but the upward forces are different (and the accelerations are not zero) …

what are the net forces? what ratio must they be?

3. Jan 24, 2014

### rude man

Four equations: two are F = ma for the man and the mass; one relates the forces on the man and the mass to each other (hint: energy conservation); and finally one relates accelerations to each other (hint: h = 1/2 a T^2).
Solve for F1 and F2 (and a1 and a2 if you want).

4. Jan 24, 2014

### rude man

That part is correct.
That part is incorrect. Acceleration is mass times the NET force on the man and the mass. One is not twice the other. Remember gravity for both sides!

5. Jan 24, 2014

### haruspex

One of the terms there is wrong by a simple factor. Correct that and you will get the answer.
No, it's quite correct.
You don't mean that.

6. Jan 24, 2014

### rude man

You're right, I don't.

I meant that, taking the man or the mass individually, each has acceleration equal to the NET force on it divided by his or its mass.
Let 1 = man
2 = mass

After actually looking at the four equations I cited it turns out that a1 = 2 a2 after all, so I missed there, but I saw no a priori reason why that should be so.

One of my sloppier posts, to be sure.

Last edited: Jan 24, 2014
7. Jan 24, 2014

### haruspex

The displacement must be in the ratio 2:1 at all times, and the time elapsed is the same for both, so the speeds and accelerations must also be in the ratio 2:1.

8. Apr 11, 2014

### Rlee03

Here is how I got it

Assume the platform with the man is 1.
Assume the platform with the block is 2.

What we know

a1=2a2
N2=2N1 (The mechanic advantage is that the man can lift twice his weight) (Equation A)

N1-m1g=-m1a1 (system with man)
N2-m2g=m2a2 (system with block)

For the man (1)

N1-m1g=-m1a1
N1-100(10)=-(100)a1
N1-1000=-100(2a2) since we know the acceleration of the man must be 2x block
N1-1000=-200a2 (Equation A)

For the system (2)

N2-m2g=m2a2 (system with block)
N2-1000=100a2
2N1-1000=100a2
N1-500=50a2
N1=50a2+500 (Equation B)
Solve Equation A and Equation B System of Equations

Plug Equation B into Equation A

50a2+500-1000=-200a2
-500=-250a2
a2=2

Plug a2 into Equation B
N1=50(2)+500=600N , which is the Normal Force on the Engineer

Plug a2 into Equation A.

N2=2N1
N2=2(600)=1200 N