View Single Post
prosteve037 is offline
Dec3-11, 09:44 PM
P: 105
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?

Phys.Org News Partner Science news on
Going nuts? Turkey looks to pistachios to heat new eco-city
Space-tested fluid flow concept advances infectious disease diagnoses
SpaceX launches supplies to space station (Update)