Torque problem (man standing on a deck)

[hri12]

Homework Statement

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.

Homework Equations

T=Frsin(Θ)

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 must be doing something wrong. Please help!

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

 

[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)

 

[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.

 

[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...

 

[HalfLostAndSearching]

First of all, it's great that you were able to solve parts 1 and 2 of the problem. For part 3, you are correct in thinking that you need to draw a free body diagram. However, there are a few things that you need to consider in order to solve for the force F.

1. The support beam is not a point, it has a certain width. This means that the force F will be distributed over a certain area on the support beam, not just at a single point. This will affect your calculations.

2. The support beam is also at an angle, so you need to take into account the component of F that is perpendicular to the beam (Fy in your diagram).

3. You also need to consider the weight of the deck and the man as external forces acting on the support beam.

With these things in mind, you can set up your equations as follows:

∑Fy = 0 (since the beam is not moving in the y-direction)
∑Fy = N + Ff - mg - Fy = 0 (since the forces are in equilibrium)
∑Fx = 0 (since the beam is not moving in the x-direction)
∑Fx = Fwall - Fx - Fw - Fm = 0 (since the forces are in equilibrium)

Now, you can use trigonometry to find the values of Fx and Fy, and then use the equation T = FrsinΘ to find the torque (T) around point A. Since you know the value of T (from part 2), you can rearrange the equation to solve for F.

F = T/(rsinΘ)

Plug in the values you have calculated and you should get the correct answer of 4580 N. Keep in mind that the coefficient of friction and normal force are not needed for this problem since there is no horizontal motion involved.

I hope this helps! Remember, don't get too frustrated and keep practicing. Science can be challenging, but with perseverance, you can definitely solve any problem that comes your way. Good luck!