Problem 7.35 (8.29)* A rope attached to a 15.0 Kg wood sled pulls the sled up a
ID: 2109351 • Letter: P
Question
Problem 7.35 (8.29)* A rope attached to a 15.0 Kg wood sled pulls the sled up a 21 degrees snow-covered hill. A 14.0 Kg wood box rides on top of the sled.Note: You will need to look up the coefficients of friction
Part A If the tension in the rope steadily increases, at what value of the tension does the box slip? Problem 7.35 (8.29)* A rope attached to a 15.0 Kg wood sled pulls the sled up a 21 degrees snow-covered hill. A 14.0 Kg wood box rides on top of the sled.
Note: You will need to look up the coefficients of friction
Part A If the tension in the rope steadily increases, at what value of the tension does the box slip? A rope attached to a 15.0 Kg wood sled pulls the sled up a 21 degrees snow-covered hill. A 14.0 Kg wood box rides on top of the sled.
Note: You will need to look up the coefficients of friction
Part A If the tension in the rope steadily increases, at what value of the tension does the box slip? Part A If the tension in the rope steadily increases, at what value of the tension does the box slip?
Explanation / Answer
Let:
M be the mass of the sled,
m be the mass of the box,
u1 be the coefficient of static friction between the box and the sled,
u2 be the coefficient of kinetic friction between the sled and the snow,
F be the friction force between the box and sled,
R1 be the normal reaction between box and sled,
R2 be the normal reaction of the snow on the sled,
g be the acceleration due to gravity,
T be the tension in the rope,
a be the acceleration of the sledge,
b the the inclination of the slope to the horizontal.
As the box is tending to slip downhill, the friction force on it is uphill.
That means the friction force of the box on the sledge is downhill.
When the box is not slipping:
mg cos(b) = R1
F - mg sin(b) = ma
F <= u1 R1
Eliminating F and R1:
a + g sin(b) <= u1 g cos(b)
a <= g[ u1 cos(b) - sin(b) ] ...(1)
For the sledge and box together:
(M + m)g cos(b) = R2
T - (M + m)g sin(b) - u2 R2 = (M + m)a
Eliminating R2:
T - (M + m)g sin(b) - u2 (M + m)g cos(b) = (M + m)a
T - (M + m)g[ sin(b) + u2 cos(b) ] = (M + m)a
T / (M + m) - g[ sin(b) + u2 cos(b) ] = a ...(2)
Combining (1) and (2), the box will slip when:
T / (M + m) - g[ sin(b) + u2 cos(b) ] > g[ u1 cos(b) - sin(b) ]
T / (M + m) > g(u1 + u2) cos(b)
T > (M + m)g(u1 + u2) cos(b).
T > (15 + 14)9.81(0.5 + 0.06) cos(21)
T > 148.64 N. to 3 sig. fig.