Two packages at UPS start sliding down the 20? ramp shown in the figure. Package
ID: 2279615 • Letter: T
Question
Two packages at UPS start sliding down the 20? ramp shown in the figure. Package A has a mass of 5.00kg and a coefficient of kinetic friction of 0.180. Package B has a mass of 10.5kg and a coefficient of kinetic friction of 0.160.
a)How long does it take package A to reach the bottom?
Explanation / Answer
Let:
m1 be the mass of package A,
u1 be the coefficient of friction between A and the plane,
a1 be the acceleration of A down the plane,
R1 be the normal reaction of the plane on A,
m2 be the mass of package B,
u2 be the coefficient of friction between B and the plane,
a2 be the acceleration of B down the plane,
R2 be the normal reaction of the plane on B,
a be the angle of the plane to the horizontal,
S be the magnitude of the reaction of each package on the other,
g be the acceleration due to gravity.
Resolve parallel and perpendicular to the plane.
For A:
m1g cos(a) = R1 ...(1)
m1g sin(a) + S - u1R1 = m1a1 ...(2)
For B:
m2g cos(a) = R2 ...(3)
m2g sin(a) - S - u1R2 = m2a2 ...(4)
Substitute for R1 from (1) in (2):
m1g [sin(a) - u1 cos(a)] + S = m1a1 ...(5)
g[sin(a) - u1 cos(a)] + [S / m1] = a1 ...(6)
Substitute for R2 from (3) in (4):
m2g [sin(a) - u2 cos(a)] - S = m2a2 ...(7)
g[sin(a) - u2 cos(a)] - [S / m2] = a2 ...(8)
Considering each package alone, and therefore putting S = 0 in (6) and (8), the fact that u1 > u2 means that a1 < a2.
The packages will therefore stay in contact, and we may equate the accelerations putting a2 = a1.
Making this substitution in (7), and adding (5) & (7):
g[ m1(sin(a) - u1 cos(a) ] + m2(sin(a) - u2 cos(a)) ] = (m1 + m2)a1
g[(m1 + m2)sin(a) - (m1u1 + m2u2)cos(a)] = (m1 + m2)a1
a1 = g[(m1 + m2)sin(a) - (m1u1 + m2u2)cos(a)] / (m1 + m2)
Substituting the numbers:
a1 = 9.81 [(6.5 + 8)sin(20) - (6.5 * 0.19 + 8 * 0.14)cos(20)] / (6.5 + 8)
= 1.86 m/s.
The time t to reach the bottom therefore satisfies:
2.0 = a1 t^2 / 2
t^2 = 4 / 1.86
t = sqrt(4 / 1.86)
= 1.47s.
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