A 12.0 m long beam is hinged to a vertical wall and held horizontally by a 5.00

Question

A 12.0 m long beam is hinged to a vertical wall and held horizontally by a 5.00 m cable attached to the wall 4.00 m above the hinge, as shown in the figure below (Figure 1) . The mass distribution of the beam is not uniform; its center of mass is at a distance of   9.80 m    from the hinge. The metal of this cable has a maximum tensile strength of 0.900 kN , which means that it will break if the tension in the cable exceeds that amount.

In this set up (where the cable is supporting the maximum mass without snapping), what is the magnitude of force with which the beam is pushing on the hinge?

Explanation / Answer

Here ,

x1 = 9.80 m

tensile strength , T = 0.900 kN = 900 N

theta = arcsin(4/5) = 53.13 degree

Now, let the mass of beam is m

T * sin(53.13 degree) * (3) - m * 9.8 * 9.80 = 0

900 * sin(53.13 degree) * (3) - m * 9.8 * 9.80 = 0

solving for m

m = 22.5 Kg

Now, for the horizontal hinge force

Fx = T * cos(theta) = 900 * cos(53.13) = 540 N

Fy = m * g - T * sin(theta) = 22.5 * 9.8 - 900 * sin(53.13)

Fy = -504.4 N

net hinge force = sqrt(504.4^2 + 540^2)

net hinge force = 740 N

the net hinge force is 740 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.