Been bothered by this equilibrium problem

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Homework Statement



The sign for Lew's Bait Shop and Sushi Restaurant is a uniform rectangle, 1.50m wide and 0.800m tall, weighing 300N. It is supported by a wire at its upper-right corner and a pivot at its lower left corner.

a) Late one night the wire was cut while the sign stays in place because the pivot has rusted tight. After the wire has been cut, find the magnitude and direction of the force and of the torque exerted by the pivot on the sign
b) When the sign was brand new, the pivot was frictionless and the wire was at the angle that minimised the tension in it. Find the tension in the wire and the magnitude and direction of the foece exerted on the sign by the pivot.

2. Homework Equations and The attempt at a solution

For a), since the sign is in equilibrium, ƩF=0. supporting force exerted by the pivot N=mg=300N upwards. Ʃτ=0, gravity exerts a torque about the pivot=F*l =300N*0.75m=225N*m clockwise, so the torque exerted by the pivot should be equal and opposite to torque exerted by the pivot. The latter part of this problem a) confuses me. There seems to be sth not right.

For b), I assumed the angle between wire and horizon is θ, so the components of T are T(x)=Tcosθ and T(y)=Tsinθ, and the components of force N exerted by the pivot are N(vertial) and N(horizontal). so T(x)=N(horizontal), T(y)+N(vertical)=mg=300N. For the torque, I was confused as well since the latter bit of problem a).

I was wondering if anyone could kindly help on this problem, please. I appreciate.
 
on Phys.org
pigiamino said:
The latter part of this problem a) confuses me. There seems to be sth not right.
Looks ok to me. What's bothering you?
For b), I assumed the angle between wire and horizon is θ, so the components of T are T(x)=Tcosθ and T(y)=Tsinθ, and the components of force N exerted by the pivot are N(vertial) and N(horizontal). so T(x)=N(horizontal), T(y)+N(vertical)=mg=300N. For the torque, I was confused as well since the latter bit of problem a).
What is the torque from the pivot now? From the wire?
 
It is easier to write up the equilibrium torque equation with respect to the pivot. The equation will contain the tension in the wire T and sine and cosine of the unknown angle θ. Consider T as function of θ. It is minimum if the derivative dT/dθ =0. You get θ from this condition.

ehild
 
haruspex said:
Looks ok to me. What's bothering you?

What is the torque from the pivot now? From the wire?

a) the second condition for equilibrium is Ʃτ about any point is 0. if there is a force exerted by the pivot on its lower left corner to the cg, and I choose cg as my point to calculate torque. I obtain a disaster that Ʃτ varies with different points I choose.

b) since a), I have no clue on b)
 
Sorry, I wasn't clear enough. I meant, what is the torque from the pivot about the pivot in part (b) of the question? What are the other torques about the pivot?
 

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