# Reaction Force

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1. Sep 23, 2015

### Jonski

1. The problem statement, all variables and given/known data
The diagram shows the instant when a long slender bar of mass 4.9 kg and length 4.9 m is horizontal. At this instant the mass m= 6.5 kg has a vertical velocity of 4.0 m/s.

If the pulley has negligible mass and all friction effects may be ignored, what is the magnitude of the reaction force on the bearing A

2. The attempt at a solution
So I split the reaction force into its x and y components Rx and Ry.

For Mass m:
T - mg = ma

For the bar:
Sum of the moments from A:
let mass of bar = B
length of bar = L
B*g*L/2 - T.L = B*L^2/3*α

Sum of forces in the x direction:
Rx = B*a=B*ω^2*L/2

Sum of forces in the y direction:
Ry - Bg+T=B*(-a) = -B*α*L/2

I solved and got Rx = 8N, however not sure how to find tension, a in the tangential, or alpha. I have tried simultaneous equations to find them but keep getting them wrong. Please help

2. Sep 23, 2015

### TSny

Hello, Jonski.

You probably need to show your attempt at solving the simultaneous equations in order for us to find your error.

Something to think about: Is the "a"in your equation T - mg = ma the same "a" as in your Rx and Ry equations?

3. Sep 26, 2015

### Tanya Sharma

Hi TSny

I get Rx = 32 N (towards right) and Ry = 37.46 N (downwards) . The magnitude of reaction force equal to 49.27 N .

Are you getting the same result ?

4. Sep 26, 2015

### TSny

Hi Tanya.
I get different values for Rx and Ry. For Rx, are you sure you're using the correct value for the speed of the center of mass of the rod?

5. Sep 26, 2015

### Tanya Sharma

Okay .

Rx = 8N (towards right) and Ry = 17.35 N (downwards) ?

6. Sep 26, 2015

### TSny

OK for Rx. But I still get a different answer for Ry.

7. Sep 26, 2015

### Tanya Sharma

I will show you my working .

8. Sep 26, 2015

### Tanya Sharma

Denoting mass of bar as M , length as L , mass of block as m ,

For block of mass m , $mg - T = ma$

Writing torque equation for the rod about the bearing, $Mg\frac{L}{2} - TL = \frac{ML^2}{3}\alpha$

$\alpha = \frac{a}{L}$

Solving this I get $T = \frac{Mmg}{2(3m-M)}$ and $\alpha = \frac{3g}{2L} - \frac{3mg}{2L(3m-M)}$ .

Now writing torque equation for the rod about the CM , $R_y\frac{L}{2} - T\frac{L}{2} = \frac{ML^2}{12}\alpha$ .

From this I get $R_y = \frac{Mg}{4} + \frac{Mmg}{4(3m-M)}$

Last edited: Sep 26, 2015
9. Sep 26, 2015

### TSny

So, positive value of $a$ implies what direction of acceleration for the block?

So, positive value for $\alpha$ implies what direction of tangential acceleration for the left end of the rod?

Does $\alpha = \frac{a}{L}$ or does $\alpha = -\frac{a}{L}$?

10. Sep 26, 2015

### Tanya Sharma

So,the only problem in post#8 is relationship between $\alpha$ and a ?

$\alpha = -\frac{a}{L}$

Last edited: Sep 26, 2015
11. Sep 26, 2015

### TSny

Yes, your first two equations in #8 look fine. I have not checked your algebra for solving for T, etc.

12. Sep 26, 2015

### TSny

Also, your equation for torque about CM looks good.

13. Sep 26, 2015

### Tanya Sharma

Are you getting $R_y = 28 N$ ?

Some doubts regarding why did we write $\alpha = -\frac{a}{L}$ ?

It is because positive value of 'a' suggests block is going downwards which in turn means the rod is rotating clockwise . But since we have considered anticlockwise positive , we write $a = -\alpha L$ .

Not sure , if this is the correct reasoning . You remember I once made an entire thread on sign issues. I am still doing some really sloppy work .

How did you think about signs in this problem?

Last edited: Sep 26, 2015
14. Sep 26, 2015

### TSny

I got 28 N too. Your reasoning for why $a$ and $\alpha$ have opposite signs is the same as how I thought about it.

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