# Mechanical Advantage and What it really means! (question inside)

1. Oct 4, 2014

### Hereformore

1. The problem statement, all variables and given/known data

The question is as follows: A man of mass 100kg is placed onto platform 1. The maximum weight that he can lift using this machine is twice his own weight (200kg*g). The mechanical advantage of the machine is constant.

If the mass of the object is 100kg, 1) what is the force upward felt on the engineer and mass respecitvely
and 2) What is their accelerations?

2. Relevant equations
Fnet=ma
MA = Fout/Fin
W=F*d

3. The attempt at a solution
So I think I'm lacking a fundamental understanding of what Mechanical Advantage is and means.

I know that the man has a downward force of 100*g N and the Normal Force is (as always) determined by his net accleration/the net force (since it's not always the exact opposite of his own weight- right?).

Lets say Man = m1
Mass = m2
Fn = Normal Force

m1g - Fn1 = -m1a1

m2g - Fn2 (mass) = m2a2

So because of the mechanical advantage applied here, if the man weighs 100g N then the output force exerted on the mass upwards is 2*100g N.

So what confuses me here is that I would have though that the normal force would then be 2*100*g Newtons for the mass, but it isnt. Is it because the system is moving? and in assuming the normal force on the mass was simply twice the weight of the man was i making the mistake of assuming normal force is anything other than a factor of the net force?

If normal force is critically a factor of net acceleration/force, then we would have, as stated above:

Fn1- m1g = m1a1

Fn2 - m2g = m2a2

But that is not enough, as we also need a relationship between acceleration, which brings me to my second point of lacking, the notion of mechanical advantage and what it means.

So I'm thinking that mechanical advantage does mean that the force out is double, but it also implies that work is conserved. So if work is conseved then:
Work man = Work mass
Fman*h= 2Fman * (1/2 h)

Is the work done by the man equal and opposite to that done by the mass? I assume it is becuse it is similar to a pulley system so they are connected so it should be (this is all intuition though id love some clarification on this).

So the man falls twice the distance as the box in the same amount of time so it's acceleration is twice that of the box.

Substituting in:

Fn1- m1g = m1a1

Fn2- m2g = m2a2 = - m2*(1/2 a1)

Further plugging in:

Since a1 = (Fn1-m1g)/m1

Fn2- m2*g = -m2 *1/2* (Fn1-m1g)/(m1))

2*Fn2 - 2m2*g = - m2 *((Fn1-m1g)/m1)

2*Fn2 = 2m2*g - m2((Fn1-m1g)/m1)

2*Fn2 = m2( 2g - Fn1/m1 + g)

2*Fn2 = m2(3g - Fn1/m1)

m1=m2=100kg

2*Fn2= 100(3g - Fn1/100)

2*Fn2 = 300g -Fn1

2Fn2 + Fn1= 300g

So this is where i get stuck, and the book i'm using says that Fnormal(mass) = 2*Fnormal(man) due to the mechanical advantage. But I thought mechanical advantage was in regards to force applied vs force out (so the weight of the man vs the normal force applied to the mass?).

If you do plug accordingly in you get the right answer:

2*(2Fnormal(man)) + Fnormal(man) = 300g

5Fnormal(man) = 300g

Fnormal(man) = 60g = 600 N

Fnormal(mass = 1200N

WOW that was long. So thanks for bearing with me. thats the correct answer, but what confuses me is that how can that ^ be the normal force when the normal force for the mass when its supposed to be double the input force? Which in this case would be 100kg*g (the weight of the man)?

The force up is not the force transmitted, i thought the force transmitted was the weight of the man unto the normal force of the mass?

Thanks for reading this far and helping out guys.

Last edited: Oct 4, 2014
2. Oct 4, 2014

### Simon Bridge

The mechanical advantage is the force amplification factor when using a tool.
If the mechanical advantage is A, then $F_{out} = AF_{in}$ ... with appropriate choices for input and output forces.
For your problem, A=2 is when the LHS is the input and A=0.5 when the RHS is the input.
LHS=input (the weight of the man) is what your book did.

Working the acceleration involves figuring the free-body diagrams for each side separately.

I wouldn't worry about "normal force" etc and just work the physics.
If the weight of the man is W, then the mass of the man is W/g and the the LHS fbd gives $W-mg/2 = Wa_{lhs}/g$ or something right?

3. Oct 4, 2014

### ehild

The man applies the normal force N1 on the platform, and it is doubled at the output, as normal force, applied to the load: N2=2N1. Imagine that he does not stand on the platform, but pushes it down with a force equal to his weight and normal to the platform - the result would be the same on the load. The normal forces are the input and the output.

m1g is the force of gravity exerted on the man. He also experiences the normal force from the platform, so the whole force acting on the man is m1g-N1, downward.
On the same way, the load experiences N2 -m2g upward force. And as you said, the work of the machine is the same on both sides, so a2=a1/2.

ehild