Answer Fun Torque Problem with 1.9m

[jcais]

Hello,

I have a homework problem. I was wondering if you all get/got the same answer that I have: 1.9m. My friend got .6m and I don't know how. I tried working it out twice and still do not come up with .6m.

Here it goes:

2 people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. if one person applies an upward force of 60 N at one end of the board, where must the other person be standing in order for the board to be in static equilibrium and with what force must they apply?

My answer:
Forces in x direction = 0
Forces in Y = F1 + F2 - W
Given:
L = 3.00 m F1 = 60 N
W = 160 N F2 = ? and x2 = ? This is x sub two.
Sum of forces and torques = 0

Sum Force = F1 + F2 - W = 0
60N + F2 - 160 N = 0
F2 = 100 N
My pivot point is at F2.
Sum of torques = 0
Torque F1 = F1(L - x2)
Torque F2 = 0 b/c at pivot point
Torque W = W(L/2 - x2)
F1L - F1x2 + (WL)/2 - Wx2 = 0
(60)(3) - 60x2 + (160 * 3)/2 - 160x2 = 0
180-60x2 + 240-160x2 = 0
420-220x2 = 0
x2 = 1.9m

Thank you for your assistance.

 

[Hootenanny]

I'm afraid I got 0.6m also.

 

[Hootenanny]

It is much easier to achieve the 0.6m answer if you take your torques about the opposite end of the board to F1. I also have F2 as 100N

-Hoot

 

[Doc Al]

Sum Force = F1 + F2 - W = 0
60N + F2 - 160 N = 0
F2 = 100 N

OK.

My pivot point is at F2.
Sum of torques = 0
Torque F1 = F1(L - x2)
Torque F2 = 0 b/c at pivot point
Torque W = W(L/2 - x2)

OK, but realize that torques have signs: if Torque F1 is clockwise, then Torque W is counter-clockwise.

F1L - F1x2 + (WL)/2 - Wx2 = 0

That's your mistake.

Hint: Try taking the other end (where F1 is) as your pivot point.

 

[jcais]

Dangit! Thank you everyone.

 

[Xamien]

I got 2.4m, which was confirmed as correct. Taking the known force as pivot point is correct but the rest of the algebra you applied was overcomplicated. I just used T_f1 + T_f2 - T_w = 0 with (F * l for each torque) and solved for l in T_f2