Tuesday, May 17, 2011
Mechanics Problems Solutions (Simple Machines: Pt. 9)
We now look at the solutions of the Mechanics problems from Part 9. The problems were as follows:
1) Show, with reasons, that the pulley system shown in Fig. 3(b) has a mechanical advantage of 2. If the weight w is lifted 1m, how much distance must the force F cover?
2) An inclined plane has an angle of Θ = π/6. The coefficient of sliding friction u = 0.3.How much force, parallel to the incline, is needed to push a 100 N object up the incline at constant speed?
3) The work done in sliding a 50 kg mass up an incline 1 m high and 5m long is 640 Joules. What is the frictional force in N?
4) In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.
5) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 m/s^2, find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done?
6)A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.
7) In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.
1) This is a single moveable pulley as with 3(a) but with just a small modification: adding another fixed pulley. Since M.A. = s/d = 2 then F = wd/ s and the machine moves a distance d while the applied force moves a distance s. Then: F = ½w. Then if the weight w is lifted 1m, the force must be applied a distance 2d = 2(1m) = 2m.
2)Note: Θ = π/6 = 30 degrees (e.g. 180 deg/ 6). We have w = mg = 100N and u = 0.3. Tnen the force F' required is: F' = mg(sin Θ + u cos Θ)
F' = (100 N) [sin 30 + 0.3 cos 30] = 100N(½ + 0.3(0.866)
F' = 100N(0.50 + 0.25)= 100N (0.75) = 75N
3)We have: mg = 50kg(10 m/s^2) = 500N, h = 1m, L = 5m, then (h/L) = 1/5 = 0.2.
But: sin Θ = h/L = 0.2 and F'.s = 640 J = F'L so F' = 640J/5m = 128N
F' = mg sin Θ + umg cos Θ so u= (F' - mg sin Θ)/ mg cos Θ
Θ = arc sin(0.2) = 11.5 deg, so cos Θ = cos (11.5) = 0.97, then:
u= (F' - mg sin Θ)/ mg cos Θ = (128 N - 500N(0.2))/ 500N(0.97)
u = (128N - 100N)/ 485 = 28N/ 485N = 0.05
Then the frictional force: F_f = uN = (0.05)(500N) =25N
4) We have r = 1 cm and R = 23 cm for the wheel and axle (Fig. 4A). The distance the load (w= mg) will move is just d = 2πr. The distance the force moves will be s = 2πR. If we take the mechanical advantage:
M.A. = s/d = (2πR)/ (2πr) = R/r = 23 cm/ 1 cm = 23
mg/ F = w/F = R/r and so: F = (r/R)w
Now, if w = 80N then F = (1/23) 80N = 3.4 N
5)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(9.8 m/s^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.
6) From the diagram of the situation (Prob. #6), we see where the man's right and left arms are. The point is that the combined vertical forces must balance a total clockwise torque or moment of 40 lbs. x 6 ft. = 240 lbs.-ft. (By the law of levers) 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 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.
7) The new set up to lift the block with less work done is shown in the diagram for Problem #7. 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.
Or: 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.