You and a friend are unloading your motorcycle off of a pickup truck using an 8f

Question

You and a friend are unloading your motorcycle off of a pickup truck using an 8ft long 2"x12" plank which has a mass of 27 kg. You slide the plank off the back end of the pickup truck so that it is horizntal with the flat ground.
(a) When very end of the plank is lying on the lip of the pickup truck bed, what force do you need to apply to the other end?
N

(b) To play with you, your friend, who has a mass of 72 kg walks out a distance 5 ft along the plank while you are holding it. What force do you need to apply now to support both the plank and your friend?
N

(c) What force does the truck apply to the other end of the plank to support the plank and your friend?
N

Explanation / Answer

Assuming that the truck doesn't move at all from the weight on it.
a) you are holding one end up and the truck is holding up the other end. The upward force of the truck and your upward force cancel out the weight of the board or else it would start to accelerate up or down due to Newton's second law, F = ma.
That is half of how to solve it. The other half is that the board isn't starting to spin faster and faster, so the net torque must also be zero. If ANY point on the board were starting to spin faster and faster, EVERY point on the board would be. This is the key to why we can pick an easy point along the board; The net torque must be zero at any point we choose, so we should choose a point that makes the equations easy.
Tau = I alpha (torque = moment of inertia times angular acceleration) is the analogous equation to F = ma. We could also say force times distance perpendicular to the force.
So lets choose the middle of the board as zero. This allows us to forget about the torque from the weight of the board.
My upward force times half the length of the board is tending to angularly accelerate the board one way.
The truck's upward force times half the length of the board is tending to angularly accelerate the board the other way.
Since there is no angular acceleration and since the distances are the same, the two forces must be equal. Call each one F
F + F = ma = 35kg*9.8m/s^2 = 2F
F = 172 N up

b) now we have another force. Up and down we have the truck's force up plus my force up equal the weight of the board plus the weight of the "friend."
F tr + F me = 9.8*(35kg + 76kg)

Now for the torque part
b) Again we can pick anywhere along the board. We can say the entire mass of the board is concentrated 4' out, and the mass of the friend is 2' out.
We don't know my force or the truck's force, so we should set zero distance at the truck or at me to get rid of one of the unknown forces. If we get rid of one of the unknown forces, we will be left with one unknown, so we will be able to solve.
Let's say the truck is at zero, then my force is tending to accelerate my end of the board up, and the two weights are tending to accelerate the board down.
Since the board isn't accelerating angularly; these torques balance one another out
F me*8' = 35kg*9.8m/s^2*4' + 76kg*9.8m/s^2*2'
F me *8 = 1372 + 1489.6 (weird units)
F me = (1372 + 1489.6)/8 N = 357.7N
c) That was the angular.
Now we use this in the up and down balance
F tr + F me = 9.8*(35kg + 76kg)
F tr = 1087.8N - 357.7N = 730.1 N
 As a check, let's look at the forces. We are each carrying half the weight of the board. The friend is three times closer to the truck than he is to me, so the truck should be carrying 3/4 of his weight and I should be carrying 1/4.

adding these two together, I have 9.8*(35/2 + 76/4) = 357.7N = F me

F tr = 9.8*(35/2 +76*3/4) = 730.1N

(this last way is using superposition)

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