What is the maximum acceleration the truck can have without the box slipping?
March 5th, 2009 | by Michael |
smiles asked:
A box of 100 kilograms of the dimensions 50 cm X 50 cm X 50 the cm is in the back of a truck-based platforms. The factors of friction between the box and the base of trucks are Î ¼ s = 0:40 and K ¼ Î = 0.20. What is the maximum acceleration the truck can have without slipping of the box?
ALI
A box of 100 kilograms of the dimensions 50 cm X 50 cm X 50 the cm is in the back of a truck-based platforms. The factors of friction between the box and the base of trucks are Î ¼ s = 0:40 and K ¼ Î = 0.20. What is the maximum acceleration the truck can have without slipping of the box?
ALI
2 Responses to “What is the maximum acceleration the truck can have without the box slipping?”
By loser on Mar 5, 2009 | Reply
Well, you do not have to worry about the coefficient of kinetic friction. You are worried about when the box is not moving relative to the bed of the truck. This is quite simple.
First, draw a free body diagram on the box. You will have the normal force from the truck bed pointing up from the bottom, and a static friction pointing in the reverse of the motion of the truck. You also have weight pointing down from the box’s center of gravity. Also, because the box is on the accelerating truck, it will accelerate at the same rate as the truck. Therefore, in the opposite direction of the friction force, note that there is acceleration. I would put the acceleration in the positive x and the friction in the negative x.
You now have a free body diagram and can sum your forces. In the x you have: fs=ma. In the y you have: N-mg=0 because the box in not accelerating in the y axis. You also have the friction equation, fs=usN.
Solve for N:
N=mg
N=100g.
Substitute this value in for N in the friction equation and solve for fs:
fs=.4(100g)
fs=40g.
Substitute this value in for fs in the summed forces in the x and solve for a:
40g=ma
40g=100a
a=2g/5 m/s^2.
You can plug whatever value in for g that your teacher wants you to plug in. I didn’t want to round.
How did you do mu?
By Rhombus on Mar 8, 2009 | Reply
Frictional force is equal to normal force x coefficient of friction. In this case, static friction (?s) is used, because the box is not in motion to begin with.
The point at which the force of the car moving forward overcomes the static frictional force of the car on the box is the point at which the box will slip off the bed. As such,
F{car} = F{normal} * ?s
dividing the mass of the box out (the box is all we are concerned with) gives us
a{car} = a{normal} * ?s
since the truck bed is flat, a{normal} = a{gravity} = 9.8m/s^2
With ?s = .40,
a{car} = 9.8 m/s^2 * .40 = 3.92 m/s^2,
which is your answer. 3.92 m/s^2.