March 22nd, 2010 | Category: zjnz.com
A 38.6 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90掳. If the beam is inclined at an angle of theta=33.4掳 with respect to horizontal, what is the horizontal component of the force exerted by the hinge on the beam? (Use the `to the right' as + for the horizontal direction.)You need to assume something about how the mass of the beam is distributed along its length, so I'm going to assume the mass is distributed "uniformly". That means the center of mass of the beam is at L/2, where L is the length of the beam and the distance from the hinge at which the cable is attached to the beam. Tension in the cable is given the symbol "T". Summing moments about the hinge and setting that summation to zero gives:
M = 0 = mgL/2cos(33.4deg)-TL
So T = mg/2cos(33.4deg) = 310.9N
Summing the horizontal forces on the beam gives this equation:
Fx = 0 = Tsin(33.4deg) - Hx (Hx is the horizontal component of the hinge force)
Therefore Hx = 171.14N
#If you have any other info about this subject , Please add it free.# |
|
edit