Two blocks are positioned on surfaces, each inclined at the same angle q = 65 o
ID: 2297007 • Letter: T
Question
Two blocks are positioned on surfaces, each inclined at the same angle q = 65o with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines as shown, so the blocks can slide together. The mass of the black block is 4 kg, and the coefficient of friction for both blocks and inclines is m = .5. Assume gravity is g = 10 m/s2.
(a) What is must be the mass of the white block if both blocks are to slide to the right at a constant velocity?
(b) What is must be the mass of the white block if both blocks are to slide to the left at a constant velocity?
(c) What is must be the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5 m/s2?
(d) What is must be the mass of the white block if both blocks are to slide to the left at an acceleration of 1.5 m/s2?
(e) Now, re-do part (d) above, but now assuming no friction at all on either incline. So in the absence of friction, what must be the mass of the white block such that both blocks slide to the left at an acceleration of 1.5 m/s2?
Would really appreciate the work done out for this problem set. Need to able to understand for my exam. Thank you very much, i really appreciate it.
Explanation / Answer
a) tension in string for black block
mgsin65 = T + 0.5*mg*cos65
T = 27.8 N
for white block
T = 0.5Mw*gcos65 +Mw*gsin65
Mw = 2.49 kg
2)in this case
tension in string for black block
T = 0.5*mg*cos65 + mgsin65
T = 44.70
for white block
mgsin65 = T + 0.5* mgcos65
m = 6.43 kg
3)in this case
tension in string for black block
mgsin65 - T - 0.5mgcos65 = ma
T = 21.8 N
for white block
T - (0.5Mw*gcos65 +Mw*gsin65) = Mw*1.5
Mw = 1.73
4)in this case
tension in string for black block
T - (0.5*mg*cos65 + mgsin65) = m*1.5
T = 50.7
for white block
mgsin65 - (T + 0.5* mgcos65) = m*1.5
m = 9.30 kg
5) tension in string for black block
T - mgsin65 = m*1.5
T = 42.25 N
for white block
mgsin65 - T = m*1.5
m = 5.58 kg
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