In this problem they show a diagram of a pulley system with an inclined plane (25 degree incline). There is a box called m1=47kg resting on the incline plane and m1=35kg hanging on the vertical line. The question reads: The coefficient of friction between m1 and the surfvace of the inclined plane are Ustatic=0.42 and Ukinetic=0.19.

Question

In this problem they show a diagram of a pulley system with an inclined plane (25 degree incline). There is a box called m1=47kg resting on the incline plane and m1=35kg hanging on the vertical line. The question reads: The coefficient of friction between m1 and the surfvace of the inclined plane are Ustatic=0.42 and Ukinetic=0.19. 
a)if the masses are held in place and then released , will they start to move? ------> I know the answer is NO. 
b)how much mass would you have to add to m2 to cause the masses to begin to move? ----> The answer should be 2.8kg but I don't know how the book got this answer. 
c)If you said no to a) and added the mass that you calculated in b), what would be the acceleration of the masses? ---> answer given in the back of textbook is 5.7 m/s^2

ANY help with this problem would be greatly appreciated!! THANK YOU SO MUCH in advance :)

Henry · Answer

W1 = m1g = 47kg * 9.8N/kg = 461 N
W2 = m2g = 35kg * 9.8N/kg = 343 N.
F1 = 461N @ 25 Deg.
Fp = 461*sin25 = 195 N.=Force parallel
to the incline.
Fv = 461*cos25 = 418 N. = Force perpendicular to incline.

Fn = F2 - Fp - Fs.
Fn = 343 - 195 - 0.42*418 = 0.
Fn = 343 - 195 - 176 = -28 N.

a.No, the masses would not move, because the opposing force is 28 N greater than F2. Therefore we must add
28 N. to F2:

b. mg = 28 N.
m = 28/g = 28 / 9.8 = 2.86 kg added.

anonymous · Answer

thank you very much! and to find the acceleration would it just be Fnet=ma and solve for a... Fnet=Fg-Ff?