(a) From static equilibrium,
(m1 + m2 + mb)g = T = 637 M
Solve for m1
(b) Now that you know all three masses, and that the weight of the beam acts through the center, do as they suggest and set the torque about either end of the beam equal to zero. That will let you solve for x. I cannot do this for you because you have not said which end has mass m1.
A beam of mass mb = 10.0 kg, is suspended from the ceiling by a single rope. It has a mass of m2 = 40.0 kg attached at one end and an unknown mass m1 attached at the other. The beam has a length of L = 3 m, it is in static equilibrium, and it is horizontal, as shown in the figure above. The tension in the rope is T = 637 N.
(a) Determine the unknown mass m1, at the left end of the beam.
(b) Determine the distance, x, from the left end of the beam to the point where the rope is attached. Note: take the torque about the left end of the beam.
3 answers
For B
left side has m1
Right side has m2
picture looks like this
----------------------------------
I T= 637
I
mb= 10kg I
------------------ ----------
<------x---------->
I I
I I
I I
m1 m2
------------L=3m-------------
left side has m1
Right side has m2
picture looks like this
----------------------------------
I T= 637
I
mb= 10kg I
------------------ ----------
<------x---------->
I I
I I
I I
m1 m2
------------L=3m-------------
the uniform beam has a mass of 180 kg under equilibrium. determine the reaction at the supports
1 answer