Applied Algebra Solutions (2)

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)
!
\!/
F

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 M_s 

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 M_s  


3m(B) = 3.2  M_s   


Then: m(B) =   3.2 M_s    / 3 m(B)   =  1.06 M_s 

Then:  m(A) =  3.2 M_s     - 1.06 M_s      = 2.14 M_s 



5) Given:   m1 + m2 = a³/P²


We need to find the period if a = 0.16 AU, and m1 + m2 =  1.8 solar masses

Write:

 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.