Show that $P=\frac{3W}{5+3\sin\theta -\cos\theta}$

Question

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I am stuck on the first part. Here is what I have done so far. First I resolved $P$ into $P\sin\theta$ and $P\cos$$\theta$. Then resolving in the horizontal plane, I obtained $F_a + P$$\cos\theta=N_b$. Then resolving vertically I obtained $P\sin\theta$ + $F_b + N_a= W$. And using the information given I obtained $2F_a=N_a$ and $2F_b=N_b$. I know I need another equation to reduce the number of variables, but i'm not quite sure how to obtain that. I tried taking moments about $A$ but wasn't able to find the perpendicular distance between $A$ and the lines of action of $P$. How do I proceed?

score 1 · Accepted Answer · answered Feb 07 '15 at 19:59

1

Compute the moments around the middle of the disc (Center of mass). The normal Forces of wall A and B will give no contribution to the resulting Moment because both lines of Action passing through the Center of mass. You will have the equation for disc radius $r$ (is the sign correct?) $F_a r + F_b r - P r = 0$ and the radius cancels out. Then you can solve a System of equations to obtain $P$.

answered Feb 07 '15 at 19:59

kryomaxim

2,882

I tried about thrice but was only able to eliminate $F_b$. $F_a$ still remains in the equation – user140161 Feb 07 '15 at 20:10
You have 5 unknown functions ($W,F_a,F_b,N_a,N_b$). Then you have a torque Balance, 2 force balances and the following equations: $2F_a = N_a, 2F_b = N_b$. Hence you can solve by $P$. – kryomaxim Feb 07 '15 at 20:14
I already know what you told me...what i'm trying to tell you is that I am still unable to solve for P – user140161 Feb 07 '15 at 20:17
was making a careless mistake...got it. – user140161 Feb 07 '15 at 21:18

Show that $P=\frac{3W}{5+3\sin\theta -\cos\theta}$

1 Answers1