Crate sliding on a flatbed truck

In summary, the force of friction between the crate and the bed is 0.30 and 0.20. The truck can have a maximum acceleration of 2.94 m/s^2 without the crate starting to slide.
  • #1
lilkrazyrae
64
0
A crate of oranges weighing 180N rests on a flatbet truck 2.0m from the back of the truck. The coefficients of friction between the crate and the bed are us=0.30 and uk=0.20. The truck drives on a straight, level highway at a constant 8.0 m/s. (a) What is the force of friction acting on the crate?
(b) If the truck speeds up with an acceleration of 1.0 m/s^2, what is the force of friction on the crate?
(C) What is the maximum acceleration the truck can have without the crate starting to slide?

Please help asap!
 
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  • #2
F=MA

[itex] \sum[/itex] F = [itex] \vec F_{fric} + \vec F_{car} [/tex]
 
  • #3
I still don't understand You can't use that equation for all three answers can you?
 
  • #4
Actually, he shouldn't be telling you much of anything, since you haven't shown how you started the problem and where you got stuck. :tongue:

Rather than tell you how to do it, I am going to ask you some questions to get you going in the right direction.

lilkrazyrae said:
(a) What is the force of friction acting on the crate?

Have you drawn a free body diagram of the crate?
What is the crate's acceleration?
What must the net force be equal to?

(b) If the truck speeds up with an acceleration of 1.0 m/s^2, what is the force of friction on the crate?

Again: Have you drawn a free body diagram?
Given the acceleration, what must the net force be equal to?
What is the maximum static frictional force?
Is that frictional force exceeded here?

(C) What is the maximum acceleration the truck can have without the crate starting to slide?

If you can get (b) then you should be able to get this.

Give it a shot and let us know what you come up with.
 
  • #5
Alright I used your questions and I got answers for all the parts but now the question is did I use your suggestions correctly. So could you tell me if I did this stuff right. On (a) I got the acceleration to be zero therefore the net force is zero. (b) F static friction equals 54N. (c) I get 54N = 180/9.8 *a which equals 2.94 m/s^2 for the max acceleration Let me know how I did please!
 

1. How does crate sliding affect the stability of a flatbed truck?

Crate sliding on a flatbed truck can affect the stability of the truck by shifting the weight distribution. This can make the truck more difficult to control and increase the risk of accidents.

2. What are the potential risks of crate sliding on a flatbed truck?

The potential risks of crate sliding on a flatbed truck include damage to the crate, damage to the truck or other vehicles, and the potential for injury to the driver or others on the road.

3. How can crate sliding be prevented on a flatbed truck?

Crate sliding can be prevented by properly securing the crate to the truck with straps or other restraints. It is also important to distribute the weight of the load evenly on the truck to prevent shifting during transport.

4. What factors can contribute to crate sliding on a flatbed truck?

There are several factors that can contribute to crate sliding on a flatbed truck, including sudden braking or acceleration, sharp turns, uneven weight distribution, and inadequate securing of the crate.

5. Are there regulations or guidelines for securing crates on flatbed trucks?

Yes, there are regulations and guidelines set by the Department of Transportation (DOT) for securing loads on flatbed trucks. These regulations include proper securing techniques, weight and size restrictions, and specific requirements for different types of cargo.

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