Two blocks are positioned on surfaces, each inclined at the same angle q =69.6 o
ID: 1490916 • Letter: T
Question
Two blocks are positioned on surfaces, each inclined at the same angle q =69.6o 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 3.7kg, and the coefficient of friction for both blocks and inclines is m = .76 . Assume gravity is g = 10 m/s2.
A)what is the mass of the white block if both blocks slide to the right at a constant velocity?
B)what is the mass of the white block, if both blocks are to slide to the left at a constant velocity?
C) what is the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5m/s2?
D) what is the mass of the white block if both blocks are to slide to the left at an acceleration of 1.5 m/s2?
e) mass of the white block with both blocks slide to the left at an acceleration of 1.5m/s2? (with no friction)
Explanation / Answer
a. if both blocks slide towards right with constant velocity
assume friction on white block = m*Mw*g*cos(69.6)
Friction on black block = m*Mb*g*cos(69.6)
Tension in rope = T
using force balance on both blocks
Mw*g*sin(69.6) = T + f = T + m*Mw*g*cos(69.6)
T = Mb*g*sin(69.6) + m*Mb*g*cos(69.6)
solving for Mw
Mw*g*sin(69.6) = Mb*g*sin(69.6) + m*Mb*g*cos(69.6) + m*Mw*g*cos(69.6)
Mw = (sin(69.6) + m*cos(69.6))*g*Mb/[g][sin(69.6 - mcos(69.6))] = (sin(69.6) + 0.76*cos(69.6))*Mb/[sin(69.6) - 0.76cos(69.6))] = 6.615622 kg
b. if both the blocks move towards black box(left), then
Mb*g*sin(69.6) - T - m*Mb*g*cos(69.6) = 0
T - m*Mw*g*cos(69.6) - Mw*g*sin(69.6) = 0
=> Mb*g*sin(69.6) - m*Mb*gcos(69.6) - m*Mw*g*cos(69.6) - Mw*g*sin(69.6) = 0
Mw = [Mb*g][sin(69.6 - m*cos(69.6))]/[g][cos(69.6) + m*sin(69.6)] = 2.0693 kg
c. if both blocks move towards right [white block] with acc = 1.5 m/s/s
Mw*g*sin(69.6) = T + f + Mw*a = T + m*Mw*g*cos(69.6) + Mw*a
T = Mb*g*sin(69.6) + m*Mb*g*cos(69.6) + Mb*a
solving for Mw
Mw*g*sin(69.6) = Mb*g*sin(69.6) + m*Mb*g*cos(69.6) + Mb*a + m*Mw*g*cos(69.6) + Mw*a
Mw = (gsin(69.6) + gm*cos(69.6) + a)*Mb/[gsin(69.6 - gmcos(69.6) - a] = (10*sin(69.6) + 10*0.76*cos(69.6) + 1.5)*Mb/[10sin(69.6) - 10*0.76cos(69.6) - 1.5)] = 9.577 kg
d. if both blocks move towards left(black box) with a = 1.5 m/s/s
Mb*g*sin(69.6) - T - m*Mb*g*cos(69.6) = Mb*a
T - m*Mw*g*cos(69.6) - Mw*g*sin(69.6) = Mw*a
=> Mb*g*sin(69.6) - m*Mb*gcos(69.6) - m*Mw*g*cos(69.6) - Mw*a - Mw*g*sin(69.6) - Mb*a = 0
Mw = [Mb][gsin(69.6 - mg*cos(69.6) - a]/[mgcos(69.6) + g*sin(69.6) + a] = 1.4293 kg
e. no friction
move to left, a = 1.5m/s/s
Mb*sin(69.6) - T = Mb*a
T - Mw*sin(69.6) = Mw*a
Mbg*sin(69.6) - Mw*a - Mwg*sin(69.6) = Mb*a
Mw = Mb(gsin(69.6) - a)/(a + gsin(69.6)) = 2.67 kg
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