# Torque Problem. Need Help

1. Dec 16, 2006

### kenau_reveas

1. The problem statement, all variables and given/known data

In a floor there is a heavy duty uniform trap door that is hinged at one side the door has a mass if M = 25 kg a length from hinge to edge of L = 1.0 m, and rotational moment of inertia I = (ML^2)/3= 8.33 kg*m^2.

Question 1:
What upward Force must be applied to handle at the edge of the door in order to hold it stationary with an angle of 60 with respect to the floor?

Question 2:
The door released and allowed to swing closed. immediately after release what is the angular acceleration of the door?

Question 3:
What angular velocity does the door have just before it slams shut?

The Attempt at a solution
for question 1:

T= r.m.g.sin(@)
= 0.5 m x 25 kg x 9.8 m/s^2 x sin 60
= 106.09 N.m

for question 2:
T= I.alfa
106.09 = 8.33 . alfa

alfa = 12.74 m/s^2

I really dont know how to do the question 3.

is this seems ok so far to you. thanks.

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Last edited: Dec 16, 2006
2. Dec 16, 2006

Please do not ignore 2. Relevant equations and 3. The attempt at a solution. Show us some effort, share your problem with us and you shall get help for sure.

3. Dec 16, 2006

### kenau_reveas

i did the changes. can you please look at it again?

4. Dec 16, 2006

For question 1, set the sum of the torques of all forces (the weight and the unknown force F) equal to zero with respect to the hinge. You can calculate the force F from this equation. It seems you didn't do the trig part correctly in part 2. Regarding question 3, you can use energy conservation.

5. Dec 16, 2006

### kenau_reveas

can you please tell me in detail what i should use in question 2 and in question 3?

6. Dec 16, 2006

Ok, let's go for question 2 first. You wrote down the equation $$T = I\cdot \alpha$$. The only torque comes from the weight. You know the moment of inertia of the door around the hinge, so you can calculate the angular acceleration easily.

7. Dec 16, 2006

### kenau_reveas

That is what i have done in the question 2. right? i know the torque and I both and then based on it i found alfa.

8. Dec 16, 2006

Yes, but as I mentioned, you missed the trig. The torque equals $$T = mg \cdot \frac{L}{2}\cdot \cos\alpha$$.

9. Dec 16, 2006

### kenau_reveas

i have used sin (@) in first one. does my first question seems right to you? and do i have to define T again in question 2?

10. Dec 19, 2006

### kenau_reveas

can anyone help me with this one?

11. Dec 19, 2006

### Hootenanny

Staff Emeritus
As radou has said above, you should use cosine and NOT sine for question one. If you are finding it difficult to visualise, then draw a triangle composed of the horizontal, the tap door and the force. Remember that you need to force and the displacement from the pivot to be perpendicular. This applies to both question one and two.