As shown in the figure, a uniform horizontal rod of mass 5kg and length 4m holds

Question

As shown in the figure, a uniform horizontal rod of mass 5kg and length 4m holds a

sign of mass 45kg. At one end, the rod is attached to the wall by a pivot; at the other end,

its supported by a cable that can withstand a maximum tension of T = 245 * SQRT(2)N

What is the height h above the pivot for anchoring the cable to the wall? (HINT: Find the angle between

the cable and the rod from the equation that sets the net torque about the pivot to zero. Then

use this angle, the length of the rod, and trigonometry to nd h. Use g = 9:8m/s^2.)

FIGURE: http://i.imgur.com/aBFfWGt.png?1

Explanation / Answer

the torques acting on the rod about its leftmost point are:

1) torque due to its own mass = 5*9.8*(2) <----as it acts at the center point(ie 4/2 = 2m)

2) torque due to the sign = 45*9.8*2

3) torque due to the tension = (force)*(distance from the leftmost point)*(sine(angle between the two forces)) <----here force = tension = 245*sqrt(2), distance = 4m , angle between the two forces = atan(h/4)

As the first two torques act in opposite dircetion to the third one , for the system to be in equilibrium,

Equating all torques,

(45*9.8)*(2)+5*9.8*2 = 245*sqrt(2)*4*sin(atan(h/4))

So, solving we get, h = 4 m <--------------------answer

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