A uniform 600 kg beam...

Question

1. A uniform 600-kg beam, 6 m long, is freely pivoted at P. The beam is supported in a horizontal position by a light strut, 5 m long, which is freely pivoted at Q and is loosely pinned to the beam at R. A load of mass M is suspended from the end of the beam at S. A maximum compression of 18,000 N in the strut is permitted, due to safety.

 

a. Under maximum load, what is the x-component of the force exerted on the beam by the pivot at P?

 

b. Under maximum load, what is the y-component of the force exerted on the beam by the pivot at P?

 

the distance between P and R is 3 m and the distance between R and S is 3 m. The distance between P and Q is 4 m. and the distance between Q and R is 5 m. P, R, and Q make a right triangle.

Answer 1

Just to verify that I've drawn the same picture you're seeing, let me describe what I'm seeing.

 

A beam and a strut are both attached to a vertical wall at points P and Q, respectively, 4 m apart. The beam extends outward horizontally 6 m. The strut angles upward 5m, and attaches to the beam at point R, which is the midpoint of the beam (3m from the wall.) The angle between the wall and the strut is θ = arctan(3m/4m). Finally, there is a mass M attached to the end of the beam at point S.

 

There are 4 forces acting on the beam, in 3 locations.

 

At point S, mass M pulls vertically downward with force W_M = Mg

 

At point R, the weight of the beam pulls vertically downward, W_B = (600kg)g

Also at point R, the strut exerts a force diagonally upward and away from the wall, with a maximum of F_R = 18,000N. The horizontal component of this force is F_Rx = 18,000 cos θ, and the vertical component is F_Ry = 18,000 sin θ.

 

At point P, the connection between the wall and the beam exerts a force (F_P) diagonally relative to the wall. It isn't immediately apparent which direction. Let us consider the two components of the force separately, F_Px and F_Py. These are the unknowns you are looking for.

 

The whole system is at equilibrium, so the net forces in each direction, and the net torque are all zero.

 

Newton's 2nd law, vertical:

F_Ry + F_Py - W_M - W_B = 0

 

There are two unknowns, W_M and F_Py. We'll come back to this.

 

Newton's 2nd law, horizontal:

F_Rx + F_Px = 0

 

F_Px = -F_Rx

Therefore, F_Px is the same magnitude as F_Rx, but pulls toward the wall, since F_Rx pushes away. Part a) complete.

 

The torques must also balance, so we need to pick an appropriate pivot point. Any of the points where force are could be a good choice.

- If you pick point S, the torque from W_M will be zero, and you can solve for F_Py.

- If you pick point P, the torque from F_Py is zero, and you could solve for W_M, then use the green equation above to solve for F_Py.

- If you pick point R, you don't have to worry about the fact that force F_R is at an angle, and there won't be any sines or cosines. You'll get a 2nd equation with two unknowns that you can solve together with the green equation again.

 

I'm going to pick point S, since we already know the vertical (perpendicular) component of F_R, and we don't really care about W_M.

 

Remember, the torque due to a force is the product of the perpendicular component of the force and the distance to the pivot point. Around point S, the torques for each force are:

τ_WM = (0m) W_M = 0

τ_R = (3m) F_Ry CCW (counterclockwise)

τ_WB = (3m) W_B CW

τ_P = (6m) F_Py

We don't know the direction of F_Py, so just pick a direction, and if it's the wrong direction, your answer will come out negative. Not a big deal, just reverse the direction you thought it was, and you're good. Let's assume up for now, which makes a CCW torque.

 

If CCW is positive The net torque is then:

τ_WM + τ_R - τ_WB + τ_P = 0

 

Your only unknown here is F_Py. You should be able to solve. Make sure you include both magnitude and direction in your answer.