Two blocks are positioned on surfaces, each inclined at the same angle theta 67.
ID: 2075217 • Letter: T
Question
Two blocks are positioned on surfaces, each inclined at the same angle theta 67.6 degrees 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 1.4 kg, and the coefficient of friction for both blocks and inclines is 1.52. Assume gravity is g=9.8 m/s2.
Picture: white block on the left of the incline and black block on the right
A) What must be the mass of the white block if both blocks are to slide to the right at a constant velocity?
B) what must be the mass of the white block if both blocks are to slide to the left at a constant velocity?
C) What 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 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 1.5m/s2?
Explanation / Answer
here,
theta = 67.6 degree
mass of black block , m1 = 1.4 kg
u = 1.52
let the mass of wite block be m2
a)
as the blocks slides to the right
net accelration , a = net force /effective mass
a = ((m1 - m2) * g * sin(theat) - u * m1 * g * cos(theta) -u * m2 * g * cos(theta))/(m1 + m2)
as accelration is zero
0 = ((1.4 - m2) * 9.81 * sin(67.6) - 1.52 * 1.4 * 9.8 * cos(67.6) - 1.52 * 9.8 * m2 * cos(67.6))
solving for m2
m2 = 0.32 kg
the mass of white block is 0.32 kg
b)
as the blocks slides to the left
net accelration , a = net force /effective mass
a = ((m2 - m1) * g * sin(theat) - u * m1 * g * cos(theta) -u * m2 * g * cos(theta))/(m1 + m2)
as accelration is zero
0 = ((m2 - 1.4) * 9.81 * sin(67.6) - 1.52 * 1.4 * 9.8 * cos(67.6) - 1.52 * 9.8 * m2 * cos(67.6))
solving for m2
m2 = 6.08 kg
the mass of white block is 6.08 kg
c)
as the blocks slides to the right with accelration , a = 1.5 m/s^2
net accelration , a = net force /effective mass
a = ((m1 - m2) * g * sin(theat) - u * m1 * g * cos(theta) -u * m2 * g * cos(theta))/(m1 + m2)
1.5 = ((1.4 - m2) * 9.81 * sin(67.6) - 1.52 * 1.4 * 9.8 * cos(67.6) - 1.52 * 9.8 * m2 * cos(67.6))/( m2 + 1.4)
solving for m2
m2 = 0.16 kg
the mass of white block is 0.16 kg
d)
as the blocks slides to the left with a = 1.5 m/s^2
net accelration , a = net force /effective mass
a = ((m2 - m1) * g * sin(theat) - u * m1 * g * cos(theta) -u * m2 * g * cos(theta))/(m1 + m2)
1.5 = ((m2 - 1.4) * 9.81 * sin(67.6) - 1.52 * 1.4 * 9.8 * cos(67.6) - 1.52 * 9.8 * m2 * cos(67.6))/(m2 + 1.4)
solving for m2
m2 = 12 kg
the mass of white block is 12 kg
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