The first problem:
The 4 kg mass is 0.2m to the left of the fulcrum. The unknown mass of the board, m, is 0.5m to the right of the fulcrum (the force acts at the center of mass of the board).
So: 4kg * 0.2m = m * 0.5m
0.8m kg = 0.5m
1.6 kg = m
The second problem:
The 2m mass is 1/4 of the length of the seesaw away from the fulcrum and rotates clockwise. Since we don't know the length of the seesaw, let's just call the length k/4, where k equals the length of the board.
The seesaw's own mass is also k/4 away from the fulcrum, but it rotates counter-clockwise.
And finally, the m mass is an unknown distance, x, away and also rotates counter-clockwise.
So putting that together:
x*m + (k/4)*(m/2) = (k/4)*2m
The masses cancel out:
x + (k/4)*(1/2) = (k/4)*2
x + k/8 = k/2
x = (3/8)k
So the mass is 3/8 of the plank's length away from the fulcrum. Since the question asks how far the mass is from the end of the plank and the fulcrum is 3/4 away from the end, subtract 3/8 from 3/4 to get 3/8.
I hope this helps. The key to both of these questions is knowing that the mass of a board/plank/etc acts at the center of mass of the object.
aside: anytime I try to write out even a small amount of math on a message board, I am reminded of how grateful I am for LaTeX.
Thank you for your answers
🙂 this is how GS solved it and I got soooo lost
Question 51
On the Surface: Under conditions of equilibrium, the force vectors in the y and x directions must sum up to 0. In this case, let's arbitrarily set the upward direction as negative and the fulcrum as the pivot point (a fulcrum is the point about which something balances). We have 3 forces at work:
the torque of the first person sitting near the fulcrum, whose torque will be given by: L = Force x lever arm (distance from the pivot point) = (2m)g x 1/4 ;
the torque of the second person sitting at a distance r: L = F x r = (mg) x (r);
the torque of the bar (in the upward direction) = F x r = -(1/2m)g x 1/4.
Note that the lever arm of the bar is 1/4 because the center of gravity of the bar is at its midpoint and the fulcrum is 3/4 of the length of the uniform bar. Summing these equations together :
Σ L = Σ Fy + Σ Fx = 0 (where Σ Fx = 0 in this case)
[(2m)g x 1/4] + [(mg) x r] + [-(1/2m)g x 1/4] = 0
2/4 mg + mgr - 1/8 mg = 0 --> 4/8 mg - 1/8 mg = -mgr --> 3/8 mg = - mgr --> r = -3/8
Therefore, the person must be 3/8 x from the bar for the system to be in equilibrium.
Going Deeper: Two rules will allow you to solve any MCAT problems dealing with torque (turning) forces: translational and rotational equilibria.
For translational equilibrium:
Σ Fx = 0 and Σ Fy = 0
The forces acting on the bar are the weights (mg). We can arbitrarily say all downward forces are positive and upward forces are negative. Substituting values for the 2 people, the bar and the upward force (xg) which maintains equilibrium at the fulcrum, we get:
Σ Fy = (2m)g + (m)g + (1/2m)g - xg = 0
Thus
x = 7/2 m
For rotational equilibrium:
Σ L = 0
The sum of the torque forces L (the force mg multiplied by the perpendicular distance from the pivot point) acting on the bar must add to zero. A uniform bar has its center of mass acting at the mid-point of the bar. Thus the line diagram is as follows:
Taking point a as our pivot point (arbitrary) and clockwise as the positive direction (arbitrary; cf. PHY 4.1.1), we get:
Σ L = (2 m) g (x) + (-7/2 m) g (3/4)x + (1/2 m) g (1/2x) + (m) g (rx) = 0
Eliminating mgx we get:
2 + (21/8) + 1/4 + r = 0
or
r = 21/8 - 2/8 - 16/8 = 3/8.
I was like
😱 
😕 WHAT!!!! I don't understand what is going on, lol. Can I use the center of mass equation as well as Torque= Force x lever arm
