ronald vellaontra
asks:
CATAGORY: Other
QUESTION: A 41 kg trapdoor, 1.2 m long, is supported by a hinge at A and a
sloping rope at B. Find the tension in the rope and the vertical and
Horizontal reactions of the hinge if the rope is just about to open the
trap door.
/
/ rope
/
1.2m /
--------------------- 65deg
A B
|
|__ Hinge
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Here the torque due to gravity just balances the torque provided by
the tension of the rope pulling at the end of the 1.2 m trap door.
The torque due to gravity about the hinge at A is given by the force
of gravity multiplied by half the length of the door since gravity
acts at the center of mass of the door.
(1) Torque(gravity) = -mg(l/2)
The torque due to the rope tension is the component of the tension at
right angles to the normal of the trap door times the length of the
door thus
(2) Torque(tension) = Tlsin(65 degrees)
the sum of these must be zero thus
(3) mg(l/2) = Tlsin(65degrees)
solving for T gives
(4) T = mg/(2sin(65degrees))
=41kg 9.8m/s^2/(2sin(65degrees))
= 221.67 N
Balancing the horizontal (x) and vertical (y) forces
(5) Tcos(65degrees) + horizontal reaction force of hinge = 0
so
(6) horizontal reaction force of hinge = -Tcos(65degrees)
= -221.67 N cos(65degrees)
= -93.68 N
In other words the hinge pulls to the left. In the vertical direction
we have
(7) Tsin(65degrees) + vertical reaction force of hinge -mg =0
or
(8) vertical reaction force of hinge = mg - Tsin(65degrees)
= 41kg 9.8m/s^2 - 221.67 N sin(65degrees)
= 401.80 N - 200.90 N
= 200.90 N
*********************************************************
* *
* T = 221.67 N *
* *
* Horizontal hinge reaction = -93.68 N (to the left) *
* *
* vertical hinge reaction = 200.90 N (directed upward) *
* *
*********************************************************
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