# Thread: Mechanics:AS level Hinge force problem!I got stuck at part c+d.

1. ## Mechanics:AS level Hinge force problem!I got stuck at part c+d.

A uniform rod AB of weight 12N is free to turn in a vertical plane about a smooth hinge at its upper end A.It is held at an angle theta to the vertical by a force P acting at B.
a)P is 5N applied horizontally.What is the force at the hinge?
b)P is horzontal and theta is arctan 3/4.What is the force at the hinge?
c)P is at right angles to AB and of magnitude 3N.What is the force at the hinge?
d)P is at right angles to AB and theta is arctan 3/4.Find P and the hinge force.

a)13N
b)12.8N
c)3underroot3 N
d)3.6N ; 10.25N

2. Hi

For c and d :

$\displaystyle \overrightarrow{R} + \overrightarrow{W} + \overrightarrow{P}= \overrightarrow{0}$

Projection on x-axis
$\displaystyle R_x - mg\sin\theta + P = 0 \Rightarrow R_x = mg\sin\theta - P$

Projection on y-axis
$\displaystyle R_y - mg\cos\theta = 0 \Rightarrow R_y = mg\cos\theta$

$\displaystyle R = \sqrt{R_x^2 + R_y^2} = \sqrt{m^2g^2 + P^2 - 2Pmg\sin\theta}$

The momentum of the 3 forces at the hinge is 0
$\displaystyle Pl - mg\frac{l}{2}\sin\theta = 0 \Rightarrow mg\sin\theta = 2P$

$\displaystyle R = \sqrt{m^2g^2 + P^2 - 2Pmg\sin\theta} = \sqrt{m^2g^2 - 3 P^2} = \sqrt{117} = 3\:\sqrt{13}$

3. yeah iam sure abt b answer
n thanks for the quick reply but i think d part is still unsolved!

what is "l " inthe last step?

4. Originally Posted by IMAN
yeah iam sure abt b answer
n thanks for the quick reply but i think d part is still unsolved!
It is quite the same as c) !

From $\displaystyle mg\sin\theta = 2P$ you can calculate $\displaystyle P = \frac12\:mg\sin\theta = 3.6 N$

$\displaystyle R = \sqrt{m^2g^2 - 3 P^2} = \sqrt{\frac{2628}{25}} = \frac65\:\sqrt{73} = 10.25 N$

5. k thanks but i still dont knwo what does "L" mean in last step of ur first reply!

6. OK

As far as I was calculating the momentum, I think that it was clear enough
I used "l" as the length of the rod

7. k .thnx.