• Jul 17th 2006, 04:06 AM
bobby77
In figure the uniform 480 kg door can swing about a horizontal axis through O. The door is shown in its equilibrium position under the control of a spring AB of stiffness 5 N/mm. How much work must be done to pull the door down to its closed position where it makes an angle of 90º to the vertical? If the door is released from its closed and latched position, what will be its angular velocity as it passes through the equilibrium position?

• Jul 17th 2006, 01:31 PM
topsquark
Quote:

Originally Posted by bobby77
In figure the uniform 480 kg door can swing about a horizontal axis through O. The door is shown in its equilibrium position under the control of a spring AB of stiffness 5 N/mm. How much work must be done to pull the door down to its closed position where it makes an angle of 90º to the vertical? If the door is released from its closed and latched position, what will be its angular velocity as it passes through the equilibrium position?

Unless I did something wrong there is a problem here.

What we need to know is by how much extra the spring has stretched in order to close the door. In order to figure this out we need to know how much the spring is stretched beyond its equilibrium position (for the spring, not the problem): I will call this distance $x_0$. Note that the total length of the spring (I'll call it D) is really the length of the "straight section" of the spring plus x0.

We can see that D is the base of an isosceles triangle with an apex angle of 30 degrees. Using any method we like we can see that D = 2.4*sin(15) = 0.621166 m or so. Now we need x0.

Sketch an extended free body diagram. There are two reaction forces Rx and Ry at the origin (which we'll be able to ignore), a weight 0.6 m along the door, and the spring force acting at an angle of 75 degrees at the end of the door. Since the door is in equilibrium we know the net torque on the door is zero. Thus by picking an axis of rotation at the origin and calling the counter-clockwise direction positive we obtain the torque version of Newton's 2nd Law to be: (L = length of the door, w = mg = weight of the door, FS is the spring force)
$\sum \tau_O = -w \frac{L}{2}sin(150^o) + F_SLsin(165^o) = 0$
(You'll need a good sketch and some angle work to get the angles. Remember the angle for the moment arm is measured from the base of the force outward from the axis.)
So the spring force comes out to be $F_S = \frac{mg}{2} \frac{sin(150)}{sin(165)} \approx 4543.72 \, N$

Now [tex]F_S = kx[tex] thus we find that x0 = 0.908743 m or so. (Remember to convert the mm in the spring constant!) This means that the unstretched (ie straight) portion of the total length of the spring is 0.621166 - 0.908743 is negative.

This means that the spring needs to stretch more than its length, which is impossible. My only thought is that the spring constant, which is usually given in units of N/cm (not mm) is incorrect.

-Dan