A uniform hinge beam 8m long weight 20N. A 25N weight hang on its free end. A rope is attached to a beam 1m from its free end to make the beam horizontal if the rope is perpendicular to the beam to ceiling. What is the tension in the rope.
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Answers
Answer:
280 N
The solution is in the picture mentioned above
Answer:
ANSWER
The beam is in equilibrium: the sum of the forces and the sum of the torques acting on it each vanish. As shown in the figure, the beam makes an angle of 60
∘
with the vertical, and the wire makes an angle of 30º with the vertical.
(a) We calculate the torques around the hinge. Their sum is 0.
TLsin30
∘
–W(L/2)sin60
∘
=0.
Here W is the force of gravity acting at the center of the beam, and T is the tension force of the wire. We solve for the tension:
T=
2sin30
∘
Wsin60
∘
=192 N
(b) Let F
h
be the horizontal component of the force exerted by the hinge and take it to be positive if the force is outward from the wall. Then, the vanishing of the horizontal component of the net force on the beam yields F
h
–Tsin30
∘
=0 or
F
h
=Tsin30
∘
=(192.3N)sin30
∘
=96.1N.
(c) Let F
v
be the vertical component of the force exerted by the hinge and take it to be positive if it is upward. Then, the vanishing of the vertical component of the net force on the beam yields :
F
v
+Tcos30
∘
–W=0 or F
v
=W−Tcos30
∘
=222N−(192.3N)cos30
∘
=55.5N