Question

A uniform beam HK of length 10m and weighting 200N is supported at both ends.A man weight 1000N stands at point p on the beam.If the reactions at H and K are respectively 400N and 800N,what is the distance HP​

Answer 1

Answer:

Dil Se Pyaar Karte Hai Dil Se Nibhayenge

Jab Tak Zinda Hai Sirf Tumhe Chahenge

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Answer 2

Explanation:

Based on my interpretation of the information, I would set the problem up like this:

x = distance HP from H to the 1000 N man

Put the 200 N weight of the beam through the center of mass at 5m (length = 10 m)

In clockwise direction, the net torque about point H must be zero to be in equilibrium:

τH = x(1000 N) + (5m)(200 N) - (10 m)(400 N) = 0

(Note the force applied at point H does not contribute to the torque about H.)

x = (10*400 - 5*200) / 1000 m = 3 m

We can check this by calculating torque about K in the same way:

(800 N)(10 m) - (200 N)(5 m) - (1000 N)(10m - x) = 0

10 - x = (800*10 -200*5) / 1000 = 7m

Then x = 10-7 m = 3m, so it checks out.