Here is the link to the EK forum explanation.

https://www.examkrackers.com/mcat/Forum/ShowThread.aspx?f=314&t=16672

I also included their explanation below just in case you can't view the thread.

I get lost right about:

"We can make the substitution that F(2) = 2*F(1) into the equation for the block: F(2) - 1000 = 100*a(2)"

I thought F(1) = 2F(2) ????? Help?

Also, does anyone really think there would be something like this on the MCAT? It seems a little extensive.

EK FORUM EXPLANATION

1. We are told that this machine can lift a block that is twice the weight of the engineer (the mechanical advantage is 2). So the machine will always work in such a way that the normal force (upward) on platform 2 is twice the normal force (upward) on platform 1.

So this means that if you place the 100kg engineer (with a weight of 1000N) on platform 1 it will exactly balance if you place a 200 kg object on platform 2 (with a weight of 2000N).

In this case the normal force on platform 1 would be 1000 N up ward and the normal force on platform 2 would be 2000 N. This would satisfy the requirement that the on platform 2 is twice the normal force on platform 1. Also, since the weight of the engineer is exactly is 1000 N and the weight of the block is exactly 2000 N, then each of the objects has its weight exactly supported by the normal force and the objects are balanced.

If you place a block with a mass greater than 200 kg on platform 2, the engineer will not be able to lift that mass, and the machine will not move. That is because the maximum amount of force the engineer can supply is his own weight (1000 N) and the machine can only multiply that by a factor of 2 to 2000 N. So if the mass on platform 2 weights more than 2000N, it cannot be lifted.

If you place a block with a mass less than 200 kg on platform 2, then the engineer's weight generates more than enough force to get the mass on platform 2 to accelerate upward.

So based on this information, we can narrow the choices down to A or B. We know the force on mass m needs to be exactly twice the normal force on the engineer.

2. In order to solve between A and B, you do need more information. Because the machine has a mechanical advantage of 2, the other result is that there is a relationship between the distances (and also the velocities and accelerations) that the two platforms move. An important general principle with simple machines is that there is an inverse relationship between force and distance. So if platform 2 has twice the force on it, it will only move half as far (as platform 1). If platform 2 only moves half as far as platform 1, that means platform 2 will always have a velocity and an acceleration that is equal to half of platform 1's.

So we know that when we put a block has a mass of less than 200 kg on platform 2, then platform 2 will accelerate up and platform 1 will accelerate down. Let's call the acceleration of platform 1 to be a(1) and the acceleration of platform 2 to be a(2). We also know that a(1) = 2*a(2) because of the paragraph above.

We also can look at the free body diagrams on the engineer and the block to find out what the forces need to be. Let's call the normal force on platform 1 to be F(1) and the normal force on platform 2 to be F(2).

The engineer has a weight of 1000 N downward and a normal force of F(1) upward. Since we know the engineer will accelerate downward, we know the 1000N weight is larger than the normal force upward. We can find the net force by subtracting the smaller one from the larger one. So the net force on the engineer is 1000 - F(1) and this is downward.

The block has a weight of 1000 N downward and a normal force of F(2) upward. Since we know the block will accelerate upward, we know the 1000N weight is less than the normal force upward. We can find the net force by subtracting the smaller one from the larger one. So the net force on the block is F(2) - 1000 and this is upward.

Now we can use Newton's second law (Fnet = m*a) with the two objects.

For the engineer:

Fnet = 1000 - F(1) = m*a(1)

1000 - F(1) = 100*a(1)

For the block:

Fnet = F(2) - 1000 = m*a(2)

F(2) - 1000 = 100*a(2)

We can make the substitution that F(2) = 2*F(1) into the equation for the block:

F(2) - 1000 = 100*a(2)

becomes

2*F(1) - 1000 = 100*a(2)

And we can also make the substitution that a(2) = 0.5 * a(1)

2*F(1) - 1000 = 100*a(2)

becomes

2*F(1) - 1000 = 100*0.5*a(1)

2*F(1) - 1000 = 50*a(1)

So finally we have an equation for the block that is:

2*F(1) - 1000 = 50*a(1)

and our equation for the engineer that is:

1000 - F(1) = 100*a(1)

This is a system of two equations and two variables, so we can solve these for what F(1) equals and what a(1) equals.

We can multiply the first equation all by 2:

2*F(1) - 1000 = 50*a(1)

becomes

4*F(1) - 2000 = 100*a(1)

and remember the second equation was

1000 - F(1) = 100*a(1)

The right hand side of each equation is 100*a(1)

So since the left hand sides of the equations are both equal to the same thing, that means the left hand sides of the equations must be equal to each other. So:

4*F(1) - 2000 = 1000 - F(1)

And we can solve for F(1)

4*F(1) = 3000 - F(1)

5*F(1) = 3000

F(1) = 600N

So this is saying the normal force on platform 1 is 600 N. This gives us choice B.