CATAGORY: Mechanics QUESTION: A sign on a rectangular board is 62 cm wide, 29cm high and has a mass of 9kg. The board can swing on an axle through the upper left hand corner, but it is held in position by a rope attached to the upper right hand corner. IF the rope leans to the left, making, forming an angle of 60 degrees with the vertical. What will be its tension? If the rope breaks allowing to the sign to swing down, what will be its initial angular acceleration? Hint: Use the parallel axis theorem to find the moment of inertia.

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The schematic figure below shows the relevant geometry.

*
|\
| \
|  \
|   \  T
|    \
|     \
|      \ a =60 degrees
|       \ |
| 62 cm  \|
hinge ->*---------*
|         |
|    *    | 29 cm
|____|____|
|
V mg = 9kg x 9.8m/s^2 = 88.2 N

For a static situation

(1) Sum of forces in the x direction = 0
Sum of forces in the y direction = 0

and/or

The total torque about the hinge = 0

We utilize the net torque = 0

(2) Torque from wire tension - torque from gravity = 0

T cos(a)L - mgL/2 = 0

where L is the length of the sign.  This comes from the fact that the
component of tension that causes rotation is T cos(a) and the lever
arm (distance to the hinge from the point of application of the force)
is the length of the sign, L, hence the torque from the tension is
Tcos(a) L.  As for the gravity part of the torque we could go into the
gory details of finding the orthognal part of the force mg to the
distance between the hinge and the center of mass but we dont have to!
It all works out to be the same as if mg were acting on any vertical
line through the center of mass so chooe the point to be at the top of
the sign in the middle.  Here the force is orthognal to the lever arm
and the lever arm is L/2.  The torque due to gravity is then -mgL/2.

Dividing (2) by L gives

(3)   T cos(a) = mg/2

T = mg/(2cos(a))

= 88.2N/(2cos(60))

= 88.2 N

If the wire breaks the initial angular acceleration (alpha) is given by

(4)    Torque(gravity) = I alpha

where I is the rotational moment of inertia of the sign about the hinge.
Solving for alpha gives

(5)  alpha = Torque(gravity)/I = (- mgL/2)/I

the moment of inertia of a rectangle about its center of mass, I(com),
is
1
(6) I(com) =  -- m (L^2 +h^2)
12

1
=  -- 9 kg((.62m)^2 +(.29m)^2)
12

= .3514 kg m^2

By the parallel axis theorem the moment of inertia with respect to the
hinge is

(7)  I(hinge) = I(com) + m l^2

where l = the distance from the com to the hinge. In this case
l^2 = (.62m)^2 +(.29m)^2 = .4685 m^2.  Thus (7) becomes

(8)  I(hinge) = .3514 kg m^2 + 9kg .4685 m^2

= 4.568 kg m^2

Substitution into (5) gives

(9)           - mgL/2        -88.2N (.62m/2)   -27.34 J
alpha = ------------ = --------------- = ------------
4.568 kg m^2    4.568 kg m^2     4.568 kg m^2