A 4.0- m-long iron bar of uniform cross section is held perpendicular to the wal

Question

A 4.0- m-long iron bar of uniform cross section is held perpendicular to the wall by a wire from the end of the bar to the wall (see the figure). The bar is fastened to the wall with a hinge. What is the tension in the wire if the iron bar weighs 250 N and the wire is 5.0 m long? Please show work.

A 4.0- m-long iron bar of uniform cross section is held perpendicular to the wall by a wire from the end of the bar to the wall. The bar is fastened to the wall with a hinge. What is the tension in the wire if the iron bar weighs 250 N and the wire is 5.0 m long? Please show work.

Explanation / Answer

In a simple way:

Ry = the vertical components of the reaction (R) at the wall
Ty = the vertical components of the tension (T) in the wire.
Since the forces are equidistant from the C.G of the bar, by taking moments about the center
Ry = Ty
Also(Ry + Ty ) = the weight of the bar = 250 N
Hence Ry = Ty = 125
Angle between the wire and rod is found from cos ? = 4/5 and hence sin ? = 3/5
T = Ty /sin ? = 125*5/3 = 208.3 N
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