We pick up now with the solutions to applications of algebra to simple machines, applications of the lever principle:
1) a) If F will move 6 times as far as the load w, then the M.A. = 6. So: F = w/6. And w = 40 kg x (9.8 ms^2) = 39.2 N. Then: F = 39.2N/6 = 6.5 N.
The work done is: W = F s = (6.5N) (10m) = 65 N-m = 65J.
(b) The mechanical advantage in this case is increased such that: Fs = 2 wd = w
2) By the principle of the lever, the combined vertical forces must balance a total clockwise torque or moment of 40 lbs. x 6 ft. = 240 lbs.-ft. The right arm exerts a counterclockwise torque at its position of 3' x F1 = 240 ft-lb. So that: F1 = (240 ft.-lbs.)/ 3 ft. = 80 lbs..
The left arm, meanwhile, exerts a second counterclockwise moment at its position (6' from the load, at the center of gravity of the plank- since it's uniform) of 6' x F2. Therefore: F2 = (240 ft.-lbs.)/(6 ft.) = 40 lbs. is the upward force exerted by his left hand.
3) Sketch is below:
-------------(4 m)-------------------^----(1m)-------x (mg)
The new set up to lift the block with less work done is shown in the diagram. This is done by lengthening the effort distance to 4 m (from 3.5 m) and shortening the load distance to 1 m (from 1.5m). Then the new force exerted needed to lift the block is found from the law of levers: F x 4m = 500N x (1m) = 500 N-m. (The 'notch' represents the pivot point, and the weigh mg is directed downwards at the end point x)
Then: F = 500 N-m/ 4m = 125 N. (Compared to the original force exerted of 214 N). The new work done is: Fs = (1/4) 500N x 1.0 m = 125 J.
4) Apply the lever principle:
m(A) + m(B) = 3.2 Ms
And : m(B) = ½ m(A)
Since m(B) is one half the mass of m(A) it must be situated at twice the distance from the center of mass, so: Bx = 2 Ax
Use the above algebraic relations, substituting m(A) = 2 m(B):
m(B) + 2 m(B) = 3.2 Ms
3m(B) = 3.2 Ms
Then: m(B) = 3.2 Ms / 3 m(B) = 1.06 Ms
Then: m(A) = 3.2 Ms - 1.06 Ms = 2.14 Ms
5) Given: m1 + m2 = a3/P2
We need to find the period if a = 0.16 AU, and m1 + m2 = 1.8 solar masses
P2 = a3/ (m1 + m2) = (0.16 )3 / (m1 + m2) = (0.00409)/ 1.8
P2 = 0.00227
Taking the square root of both sides:
P = 0.048 yrs.