# Torque problem (man standing on a deck)

1. Apr 27, 2014

### hri12

1. The problem statement, all variables and given/known data
A man stands on the edge of a deck of width W = 4.8 m. His mass is m1 = 73 kg, and the deck's mass is m2 = 600 kg, uniformly distributed over its surface. The deck is held in place by a support beam which is forming and angle theta = 37 deg with the side of the house.

.......\o/
......../\
|-----
|...../ F
|..../
|.../
|../
|./
|/ <= 37

1. Calculate the torque around point A due to the weight of the man.
2. Calculate the torque around point A due to the weight of the deck.
3. Calculate the force F that the deck is exerting on the support beam.

I found the answers to 1 and 2 pretty easily (I posted them in case of relevance), I cannot for the life of me figure out 3 though. The answer is supposed to be 4580 N. The closest I get is 4880.

2. Relevant equations

T=Frsin(Θ)

3. The attempt at a solution

1. T = FrsinΘ
T = marsinΘ
T = 3433.92 Nm

2. T = FrsinΘ
T = ma(r/2)sinΘ
T = 14112 Nm

3. Draw the free body diagram. This is where I start to have problems.

.......\o/
..(w) /\
|-----
|...../ F
|..../
|.../
|../
|./
|/ <= 37

First, I want the hinge to be at the upper left hand corner. So, I know that the wall exerts a force in x (w), I know that there is a force of friction between the wall and the deck/ plank. I know that the force F needs to be divided in x and y. I know that the center of mass of the plank is in the middle, which is where there is the force of the weight and I think (I'm not sure) that there should be a normal force from where the support beam, well, supports the deck.

If I am correct, then:

∑Fy = 0
∑Fy = N + Ff - mg - Fy
and
∑Fx = 0
∑Fx = Fwall - Fx.

There are too many unknown variables though. I don't have the coefficient of kinetic friction, I don't know how to calculate the normal force and I have no values for the force of the wall.

I'm getting that frustrated knot in my stomach. I'm into my third hour of just this problem now.

2. Apr 27, 2014

### hri12

I think that I might need to use Pythagoras at some point.

I know that Fy= F sin(theta) and Fx = F cos(theta)

3. Apr 28, 2014

### PhanthomJay

You first need to determine the horizonatal support reaction at the deck/wall connection by summing moments about the base. That horiz force gets transmitted internally to the deck/beam joint. The vert force at that same joint is the man's weight and half the deck weight. Now use Pythagoras.

Note that it appears that the problem is looking for the force in the slanted 'beam' under both the deck and man loading, not just the deck.

You've got an extra force in your sum of forces in y direction equation (Fy or N?) . Use a free body diagram of the deck, assumed pinned at each end. And mg of both man and deck are required, as I check the given answer. I don't know what you mean by kinetic friction force. The deck is anchored to the wall by mechanical means, subject to vert and horiz forces at that joint.

4. Apr 28, 2014

### BvU

Please show us how you find the 4880

You know the supporting beam does not rotate around A (which I assume is where the <== 37 points ?).
So the net torque must be zero....