Statics/Equilibrium - Find the Tension in the Wire

prosteve037

1. The problem statement, all variables and given/known data
I've already solved the problem, though I didn't understand WHY I took the steps I did; I just want to know why this is the way to solving the problem. Here's the question:

In the figure below, a thin horizontal bar AB of negligible weight and length L = 1.9 m is hinged to a vertical wall at A and supported at B by a thin wire BC that makes an angle θ = 42° with the horizontal. A block of weight W = 140 N can be moved anywhere along the bar; its position is defined by the distance x = 1.06 m from the wall to its center of mass. Find (a) the tension in the wire, and the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A


2. Relevant equations
[itex]\sum{F}_{x}\textit{ = ma}_{x}\textit{ = 0}[/itex]

[itex]\sum{F}_{y}\textit{ = ma}_{y}\textit{ = 0}[/itex]

[itex]\sum{\tau}\textit{ = 0}[/itex]

3. The attempt at a solution

a.) Tension in the wire, T :

[itex]\sum{\tau}\textit{ = 0 = TsinθL - W}_{block}\textit{x}[/itex]

[itex]\textit{T = }\frac{W_{block}x}{sinθL}[/itex]

[itex]\textit{T = }\frac{(140 N)(1.06 m)}{sin(42°)(1.9 m)}[/itex]

[itex]\textit{T = 116.7265 N}[/itex]


Now b.) and c.) can be found easily using the other force relations.


My question is, why can't you use the force relations instead of the torque relation? I can see that it gives the wrong answer, but WHY does it give the wrong answer if you don't use the torque relation?

Is it because the weight of the bar is neglected? If so, how does that affect what you can or can't use in a problem like this?

Thanks

PhanthomJay

It doesn't give the wrong answer. To solve part b and c, you use the force relationship. When you look at the system, you have 3 unknown external reaction forces..the x and y forces at A, and the force at B directed along the wire..so you need 3 equations to solve the problem. Don't forget the reaction at B.....maybe that's what you forgot when you say you got the wrong answer. The vert force at B plus the vert force at A must equal W. The horiz force at A is balanced by the horiz force at B.

prosteve037

Just realized I forgot to provide the figure :P Here it is:





Ohhh so then I think I set up the force equations wrong. I think I forgot to include the components of the contact force from the wall, thinking that they were just negligible 3rd Law force pairs :P


So then here are my 2 revised force equations for this problem:

[itex]\textit{F}_{y}\textit{ + Tsinθ = W}_{block}[/itex]

[itex]\textit{Tcosθ = F}_{x}[/itex]

Look good?


I keep making the elementary mistake of excluding forces that are acting on the body of interest, thinking that Newton's 3rd Law would cancel them out. That said, should I just always focus on the forces acting ON a body and disregard the 3rd Law? That is, for problems where the system as a whole isn't accelerating?

PhanthomJay

Yes, which direction?yes, which direction? yes I wouldn't disregard it completely, it comes in handy when looking at forces on different bodies For accelerating bodies, the process is the same: look at the forces acting on the body. But Newton's 3rd Law applies to accelerating bodies as well as to non- accelerating bodies.