https://session.masteringphysics.com/problemAsset/2035666/1/1026453.jpg A unifor

Question

https://session.masteringphysics.com/problemAsset/2035666/1/1026453.jpg A uniform beam 3.85 m long and weighing 3000 N carries a 3700 N weight 1.50 m from the far end, as shown in the figure below (Figure 1) . It is supported horizontally by a hinge at the wall and a metal wire at the far end. Part A How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? T = N SubmitMy AnswersGive Up Part B What are the horizontal component of the force that the hinge exerts on the beam? Fh = N SubmitMy AnswersGive Up Part C What are the vertical component of the force that the hinge exerts on the beam? Fv = N

Explanation / Answer

The tension on the string can be spit into two components as Tsin(30) along +ve y axis and T cos(30) along -ve x-axis.The weight of the beam acts on the center of mass, ie at 3.85/2 m from the left end.

a)
The net torque about the hinge
= 3000 x 3.85/2 + 3700 x (3.85 - 1.5) - T sin(30) = 0
14470 = 0.5 T
T = 28940 N

b)
The horizontal component of of the hinge is balanced by the T cos(30) component of the string.
Fh = T cos(30)
= 28940 x cos(30)
= 25062.775 N

c)
The net vertical force can be written as
Fv - 3000 - 3700 + T sin(30) = 0
Fv = T sin(30) - 6700
= 14470 - 6700
= 7770 N

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