(adsbygoogle = window.adsbygoogle || []).push({}); 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 barABof negligible weight and lengthL= 1.9 m is hinged to a vertical wall atAand supported atBby a thin wireBCthat makes an angleθ= 42° with the horizontal. A block of weightW= 140 N can be moved anywhere along the bar; its position is defined by the distancex= 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 atA

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

**Physics Forums - The Fusion of Science and Community**

# Statics/Equilibrium - Find the Tension in the Wire

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Statics/Equilibrium - Find the Tension in the Wire

Loading...

**Physics Forums - The Fusion of Science and Community**