How Do You Calculate the Weight on a Suspended Board in Static Equilibrium?

[akan]

Homework Statement

A uniform board of length L and weight W is suspended between two vertical walls by ropes of length L/2 each. When a weight w is placed on the left end of the board, it assumes the configuration shown in the figure.

http://img260.imageshack.us/img260/3992/rw1234ok6.jpg

Find the weight w in terms of the board weight W.

Homework Equations

If the left rope corresponds to T_1, and the right rope to T_2, left and down is negative, right and up is positive, then:

Sum(F_x) = -T_1 sin(35) + T_2 sin(60) = 0
Sum(F_y) = T_1 cos(35) + T_2 cos (60) - W - w = 0
Sum(T_z) = - W(L/2) cos(9.2) + T_2 L sin(20.8 [just calculated this from graphical drawing]) = 0 [pivot at w].

The Attempt at a Solution

eq1. T_2 sin(60) = T_1 sin(35)
eq1. T_2 = T_1 sin(35)/sin(60)

eq2. T_1 (cos(35) + sin(35)cos(60)/sin(60)) - W = w

eq3. T_2 L sin(20.8) = W (L/2) cos(9.2)
eq3. T_2 sin(20.8) = W (1/2) cos(9.2)
eq3. T_1 sin(35)sin(20.8)/sin(60) = W (1/2) cos(9.2)
eq3. T_1 = [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)]

eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W * .4274425413] / [.203680986] * [1.150306554] = w
eq2 w = 2.41 W

The correct answer is 1.42 W. What did I do wrong?

 

[tiny-tim]

eq2. [W (1/2) cos(9.2) sin(60)] / [sin(35)sin(20.8)] * [(cos(35) + sin(35)cos(60)/sin(60))] - W = w
eq2. [W * .4274425413] / [.203680986] * [1.150306554] = w
eq2 w = 2.41 W

Hi akan!

You left out the - W.

(though it might have been quicker

i] to find T1 in terms of T2 instead of vice versa

ii] to take moments about the point where the lines of the two ropes meet )

 

[akan]

Hi akan!

You left out the - W.

(though it might have been quicker

i] to find T1 in terms of T2 instead of vice versa

ii] to take moments about the point where the lines of the two ropes meet )

Thank you!

1) Yes, you're right, that's easier.
2) Uh... If I did that, how would I account for the torque due to W and w? Where would the moment arm be? Wouldn't that make more unknown variables, too?

 

[tiny-tim]

2) Uh... If I did that, how would I account for the torque due to W and w? Where would the moment arm be? Wouldn't that make more unknown variables, too?

Well, you know that the combined torque of W and w must go through that point … and they're both vertical, so you just balance the left and right amounts, like an uneven see-saw.

(you get the same equation in the end, of course …*but it's all in one geometrical diagram, so it's easier to check what you're doing)

 

[akan]

Can you please explain how to express torque if I set the pivot point in such a way that it does not even lie along the level arm? This is something that neither my textbook nor my professor seem to be able to explain clearly. Thanks.

 

[tiny-tim]

Can you please explain how to express torque if I set the pivot point in such a way that it does not even lie along the level arm? This is something that neither my textbook nor my professor seem to be able to explain clearly. Thanks.

(btw, the lever arm (stupid name, IMO) isn't the line of the force … it's the distance from the pivot point to the line of the force).

You can set the pivot point anywhere

(The pivot point I suggested has the advantage that it makes two of the lever arms (and therefore the torques) zero! )

torque = force times perpendicular distance …

that is, the perpendicular distance (shortest distance) from the pivot point to the line of application of the force.

This extract from the PF Library may help:

Every force F has a Moment about any point P.

To find the Moment, draw R, the point of application of the force, and L, the line of application of the force, and draw the perpendicular line PQ from P to L (so both Q and R lie on L).

For the Moment of a velocity, R is the position of the centre of mass, and L is the line of the velocity.

Then the Moment of F about P is the vector written "r x F" (pronounced "r cross F"), where r is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

r is sometimes called the lever arm.

Note that if P is on the line L, then P = Q, so PQ = 0, so the Moment of the force is 0.

In nearly all exam problems, everything is in the same plane (the plane of the examination paper!), so all the Moment vectors are vertically out of the page.

In other words, they're all parallel to each other, so we can forget that they're vectors, and treat them simply as numbers, F times PQ.

We can take Moments about any point, so we always choose whatever point makes the calculations easiest.

Usually, it's the point of application of an unknown force, so that the Moment of that force is 0, making the equation shorter!

Any questions?

 

[akan]

http://img228.imageshack.us/img228/5301/37776389yh0.png
http://g.imageshack.us/img228/37776389yh0.png/1/

Ok, is this what you meant?

If yes, how do I find the distance from W_CM to PIVOT, and the angle that its arm makes with the horizontal? Thanks.

 

[tiny-tim]

Hi akan!

Ok, is this what you meant?

Yes!

If yes, how do I find the distance from W_CM to PIVOT, and the angle that its arm makes with the horizontal? Thanks.

You only need the horizontal distances from P (the pivot) to W and w … the forces w and W are vertical, so the "lever arm" (horrible phrase) is horizontal.

The horizontal distance to w is z₁sin35º, and the horizontal distance to W is (L/2)cos9.2º minus that.

 

[akan]

Shouldn't the distances be perpendicular to the lever arm (please also suggest a better term), because torque is r x F = rF sin theta, where theta is the angle between the two vectors? And please explain how'd you get the horizontal distance to W from the pivot. Thanks.

 

[tiny-tim]

Shouldn't the distances be perpendicular to the lever arm (please also suggest a better term), because torque is r x F = rF sin theta, where theta is the angle between the two vectors? And please explain how'd you get the horizontal distance to W from the pivot. Thanks.

The distance is the lever arm.

In r x F = rF sin theta, F is the force (along the "line of force"), and r is the lever arm.

I just call r the "perpendicular distance".

The horizontal distance to w is z₁sin35º (and you can get z₁ from the sine rule), and the horizontal distance to W is (L/2)cos9.2º minus that.

 

[akan]

Can you please draw a picture from which it would be clear where the (L/2) cos 9.2 comes from? Thank you.

 

[tiny-tim]

Can you please draw a picture from which it would be clear where the (L/2) cos 9.2 comes from? Thank you.

erm … it's your picture …

(L/2) cos 9.2º is the horizontal distance from W (the centre of the board) to w.

 

[akan]

Ah. I see what you're saying. But I need the horizontal distance from W to the pivot, not from W to w.

 

[tiny-tim]

grrr!

The horizontal distance to w is z₁sin35º (and you can get z₁ from the sine rule), and the horizontal distance to W is (L/2)cos9.2º minus that.