Reaction Force, find tension, a in the tangential, or alpha

[Jonski]

Homework Statement

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

 

[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?

 

[Tanya Sharma]

Hi TSny

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

Are you getting the same result ?

 

[TSny]

Hi TSny

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

Are you getting the same result ?

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

 

[Tanya Sharma]

Okay .

R_x = 8N (towards right) and R_y = 17.35 N (downwards) ?

 

[TSny]

Okay .

R_x = 8N (towards right) and R_y = 17.35 N (downwards) ?

OK for R_x. But I still get a different answer for R_y.

 

[Tanya Sharma]

I will show you my working .

 

[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 $R_y = \frac{Mg}{4} + \frac{Mmg}{4(3m-M)}$

 

[TSny]

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

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

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

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

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

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

 

[Tanya Sharma]

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

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

 

[TSny]

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

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

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

 

[TSny]

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

Also, your equation for torque about CM looks good.

 

[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?

 

[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.