Two blocks are positioned on surfaces, each inclined at the same angle q =60.6 o
ID: 1375848 • Letter: T
Question
Two blocks are positioned on surfaces, each inclined at the same angle q =60.6o 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 3.5 kg, and the coefficient of friction for both blocks and inclines is m = 1.03. Assume gravity is g = 10 m/s2.
What is must be the mass of the white block if both blocks are to slide to the right at a constant velocity?
What is must be the mass of the white block if both blocks are to slide to the left at a constant velocity?
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?
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?
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
given,
mass of black box = 3.5 kg
angle of inclination = 60.6 degree
coefficient of friction = 1.03
if both the blocks are moving to the right side with constant velocity then
suppose black box is on left side as figure is not visible in question
for white box
Mwg * sin(60.6) - kMwg * cos(60.6) - T = Mwa
constant velocity then a = 0
Mwg * sin(60.6) - kMwg * cos(60.6) - T = 0
for black box
T - Mbg * sin(60.6) - kMbg * cos(60.6) = Mba
constant velocity then a = 0
T - Mbg * sin(60.6) - kMbg * cos(60.6) = 0
equating both the equations
Mbg * sin(60.6) + kMbg * cos(60.6) = Mwg * sin(60.6) - kMwg * cos(60.6)
3.5 * 10 * sin(60.6) + 1.03 * 3.5 * 10 * cos(60.6) = Mw * 10 * sin(60.6) - 1.03 * Mw * 10 * cos(60.6)
Mw = 13.1816 kg
mass of the white block if both blocks are to slide to the right at a constant velocity = 13.1816 kg
if box is moving on left side
Mbg * sin(60.6) - kMbg * cos(60.6) = Mwg * sin(60.6) + kMwg * cos(60.6)
3.5 * 10 * sin(60.6) - 1.03 * 3.5 * 10 * cos(60.6) = Mw * 10 * sin(60.6) + 1.03 * Mw * 10 * cos(60.6)
Mw = 0.9293 kg
mass of the white block if both blocks are to slide to the left at a constant velocity = 0.9293 kg
if there is acceleration of 1.5 m/s^2 then
if the white block is moving on right side then,
Mbg * sin(60.6) + kMbg * cos(60.6) + Mb * a = Mwg * sin(60.6) - kMwg * cos(60.6) - Mw * a
3.5 * 10 * sin(60.6) + 1.03 * 3.5 * 10 * cos(60.6) + 3.5 * 1.5 = Mw * 10 * sin(60.6) - 1.03 * Mw * 10 * cos(60.6) - Mw * 1.5
Mw = 24.7884 kg
mass of the white block if both blocks are to slide to the right at an acceleration of 1.5 m/s2 = 24.7884 kg
if the white block is moving on left side then,
Mbg * sin(60.6) - kMbg * cos(60.6) - Mb * a = Mwg * sin(60.6) + kMwg * cos(60.6) + Mw * a
3.5 * 10 * sin(60.6) - 1.03 * 3.5 * 10 * cos(60.6) - 3.5 * 1.5 = Mw * 10 * sin(60.6) + 1.03 * Mw * 10 * cos(60.6) + Mw * 1.5
Mw = 0.49418 kg
the mass of the white block if both blocks are to slide to the left at an acceleration of 1.5 m/s2 = 0.49418 kg
if friction force is 0 then
Mbg * sin(60.6) - Mb * a = Mwg * sin(60.6) + Mw * a
3.5 * 10 * sin(60.6) - 3.5 * 1.5 = Mw * 10 * sin(60.6) + Mw * 1.5
Mw = 2.4718 kg
mass of the white block such that both blocks slide on frictionless surface to the left at an acceleration of 1.5 m/s2 = 2.4718 kg
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