Hanging sign, torque and pivot points

[Sean77771]

Homework Statement

There is a sign of mass M that is attached to a rigid bar perpendicular to the ground. There is also a rope attached to the sign at the same point, then pulled out at an angle 45 degrees below the horizontal up to the ceiling.

a) Consider the rigid bar to be at equilibrium while the sign is hanging. What are the net force (in Newtons) and the net torque (in N m) acting on the bar?

b) Identify all forces that act on the bar. Write out an equation for the net force on the bar.

c) Choose the point where the bar meets the wall as a pivot point. Write out an equation for the net torque on the bar about that point.

d) Determine the magnitude of the tension in the rope. Take the mass of the sign to be 25kg, the mass of the bar is 5kg, and the length of the bar is 2m.

e) We could choose any point as the pivot and still get the same net torque. Why was the choice in part c a good choice for the pivot?

Homework Equations

T_net = Ia
when in equilibrium,
-- F_net = 0
-- T_net = 0

The Attempt at a Solution

I got that, for (a), these are both equal to zero, as the bar is in equilibrium...don't know where to go from here.

 

[Shooting Star]

If you give a rough diagram, I'm sure you'll get some responses. I am unable to visualize the whole picture.

Anyway, in all these kinds of statics problems, the way to proceed is to equate the sum of the horizontal and vertical forces individually to zero; and to equate the moment of all the forces about a conveniently chosen point to zero. Then solve the resulting equations.

 

[ogb p]

For (b), the forces acting on the bar are the weight of the sign (Fg = mg), the tension in the rope (T), and the normal force from the pivot point (Fn). The equation for the net force would be F_net = Fg + T + Fn = 0.

For (c), the equation for the net torque would be T_net = r x F = 0, where r is the distance from the pivot point to the point where the rope is attached to the sign.

For (d), to determine the tension in the rope, we can use the equation T_net = Ia. Since the bar is not rotating, the net torque is equal to zero, so T_net = 0 = r x T - mgL/2, where r is the distance from the pivot point to the center of mass of the sign, L is the length of the bar, and a is the angular acceleration, which is also equal to zero. Solving for T, we get T = mgL/2r = (25kg)(9.8m/s^2)(2m)/(2m) = 245N.

For (e), the choice of the pivot point in part (c) was a good choice because it allowed us to eliminate the torque term in the equation and solve for the tension in the rope directly. Choosing a different pivot point would introduce additional variables and make the equation more complicated. Additionally, choosing a pivot point that is not at the point where the bar meets the wall would result in a non-zero net torque, which would mean the bar is not in equilibrium.