A uniform beam 4.50 m long and weighing 2150 N carries a 3550 N weight 1.50 m fr

Question

A uniform beam 4.50 m long and weighing 2150 N carries a 3550 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. How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? What are the horizontal component of the force that the hinge exerts on the beam? What are the vertical component of the force that the hinge exerts on the beam?

Explanation / Answer

1. The net horizontal forces on the beam are the horizontal component of the hinge, H(x), and the horizontal component of the tension in the cable, Tcos:

F(x) = 0 = H(x) - Tcos
H(x) = Tcos---->(1)

The vertical forces (net) are:

F(y) = 0 = H(y) + Tsin - 2150N - 3550N
H(y) = 5700N - Tsin------>(2)


Since we have three unknowns, the hinge force in twodirections and the tension, we need one more equation. The problem satisfies the two conditions for equilibrium (the two force equations are one of those conditions, the other is torque). The equation for the net torque is:

= 0 = (H) + (T) + (beam) + (weight)

The torque around the hinge (H) is zero because it has a zero moment arm and the equation for torque is = rFsin, therefore:

0 = rTsin - rw - rW
T = (rw + rW) / rsin
= [(2.00m(2150N) + 2.50m(3550N)] / (4.00m)sin30.0°
T= 6587.5N

This is the minimum tension the cable needs to be able to have to support the beam and weight.


2. The horizontal tension is found from (1) now that we know T:

H(x) = Tcos
= (6587.5N)cos30.0°
= 5704.9 N

3. The vertical from (2):

H(y) = 5700N - Tsin
= 5700N - (6587.5N)sin30.0°
= 2406.25N

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