# Torque problem (man standing on a deck)

**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 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.

http://ift.tt/1isq8sV

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