Posted on June 1, 2010.
What is the maximum acceleration the truck can have without the box slipping? A 100 kg box of dimensions 50 cm x 50 cm x 50 cm at the back of a flatbed truck. The coefficients of friction between the casing and the chassis of the truck are a¬s = 0.40 and μk = 0.20. What is the maximum acceleration the truck can have without the box slipping?
Well, you do not have to worry about the coefficient of kinetic friction. You're worried when the box does not move relative to the truck bed. It's very simple.
First, draw a free body diagram on the box. You will have the normal force from the truck bed up from the bottom, and a static friction pointing in the opposite direction of movement of the truck. You also have the weight down the center of gravity zone. Also, because the box is on the truck accelerates, it will accelerate at the same speed as the truck. Therefore, in the opposite direction of the friction force, note that there is an acceleration. I would put the acceleration of the positive x and negative friction in the x
You now have a free body diagram and can summarize your strengths. In the x you: fs = ma. In there you have: N-mg = 0 because the box does not accelerate in the y-axis You also have the equation of friction, fs = USN.
Solve for N:
N = mg
N = 100g.
Substitute this value in the N in the equation to solve friction and FS:
fs =. 4 (100g)
fs = 40g.
Substitute this value in fs forces summed in the x and solve a:
40g = ma
40g = 100
a = 2g / 5 m / s ^ 2.
You can connect any value g that your teacher asks you to plug in, I did wrong.
How did you mu?
friction force is equal to the normal force x coefficient of friction. In this case, the friction (ms) is used, because the box is not moving initially.
The point where the force of the car go forward overcomes the force of static friction of the car on the box is at the point where the box slides out of bed. As such,
F () = F car (normal) * a¬s
dividing the mass of the outlet box (the box is what we are concerned) gives us
a car () = (a * a¬s normal)
from the truck bed is flat, a normal () = () density = 9.8 m / s ^ 2
With a¬s = .40,
a car () = 9.8 m / s ^ 2 * 0.40 = 3.92 m / s ^ 2
What is your answer. 3.92 m / s ^ 2.
