Finding maximum acceleration of truck without having package fall off.

In summary: Thanks!In summary, the truck can have a maximum acceleration of 4.0 m/s² without the package falling off the back.
  • #1
soul_observer
10
0

Homework Statement


Anthony is going to drive a flat-bed truck up a hill that makes an angle of 10 degrees wtih respect to the horizontal direction. A 36.0-kg package sits in the back of the truck. The coefficient of static friction between the package and the truck bed is 0.380. What is the maximum acceleration the truck can have without the package falling off the back?


Homework Equations


Force of friction<(=) mu of static friction(normal force)


The Attempt at a Solution


So I drew a FBD. I found two forces in the horizontal: the force of friction and the force of the pull. And two vertical forces: the normal force and the force of gravity. I aligned the Normal force, the force of friction and the force of the pull with the axis, and then split the gravity into x and y components.

4. Confusion
Am I correct on the horizontal force of pull? And would this be acceleration? If so, is it appropriate to say that the sum of forces equals the force of friction + the force of acceleration = ma? This is where I'm confused. I think that once I clear this up I can solve the problem. Thanks for any help!
 
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  • #2
soul_observer said:

Homework Statement


Anthony is going to drive a flat-bed truck up a hill that makes an angle of 10 degrees wtih respect to the horizontal direction. A 36.0-kg package sits in the back of the truck. The coefficient of static friction between the package and the truck bed is 0.380. What is the maximum acceleration the truck can have without the package falling off the back?


Homework Equations


Force of friction<(=) mu of static friction(normal force)


The Attempt at a Solution


So I drew a FBD. I found two forces in the horizontal: the force of friction and the force of the pull. And two vertical forces: the normal force and the force of gravity. I aligned the Normal force, the force of friction and the force of the pull with the axis, and then split the gravity into x and y components.

4. Confusion
Am I correct on the horizontal force of pull? And would this be acceleration? If so, is it appropriate to say that the sum of forces equals the force of friction + the force of acceleration = ma? This is where I'm confused. I think that once I clear this up I can solve the problem. Thanks for any help!
I believe you are confused about the 2 forces acting on the package in the x direction. There are two forces in the x direction (parallel to the incline): The gravity force acting down the plane and the friction force acting up the plane. Now use Newton 2 to solve for 'a'_max. Do you see why the friction force acts up the plane?
 
  • #3
soul_observer said:

Homework Statement


Anthony is going to drive a flat-bed truck up a hill that makes an angle of 10 degrees wtih respect to the horizontal direction. A 36.0-kg package sits in the back of the truck. The coefficient of static friction between the package and the truck bed is 0.380. What is the maximum acceleration the truck can have without the package falling off the back?


Homework Equations


Force of friction<(=) mu of static friction(normal force)


The Attempt at a Solution


So I drew a FBD. I found two forces in the horizontal: the force of friction and the force of the pull. And two vertical forces: the normal force and the force of gravity. I aligned the Normal force, the force of friction and the force of the pull with the axis, and then split the gravity into x and y components.

4. Confusion
Am I correct on the horizontal force of pull? And would this be acceleration? If so, is it appropriate to say that the sum of forces equals the force of friction + the force of acceleration = ma? This is where I'm confused. I think that once I clear this up I can solve the problem. Thanks for any help!

Please correct me if I'm misunderstanding, but I visualize the X axis is aligned with the surface of the truck bed and the Y axis 90º to the bed?

So looking at a FBD of the package will have three basic forces - the force of gravity (both an X and a Y component), the force of friction (x component only), and the inertial force opposite to the force of the motion of the truck (x component only).

Since the package cannot move in the Y axis, we would only look at the x components. If the sum of the X components = 0, the package will be on the verge of slipping. So Forcefriction + Forcegravity, x-component + Forceinertia = 0

Please let me know if I missed anything or made anything more confusing.
 
  • #4
Yes I believe I understand, because the frictional forces are perpendicular to the area of contact? Ahhh, this is always the problems I have with physics! :) Thanks much for the help.
-Jasmine
 
  • #5
So is the Force of inertia the force you are solving in this problem? Would that be the acceleration? I am a bit confused.
 
  • #6
soul_observer said:
So is the Force of inertia the force you are solving in this problem? Would that be the acceleration? I am a bit confused.

Once you find the force of inertia you can use F = ma to find acceleration.
 
  • #7
Stovebolt said:
Once you find the force of inertia you can use F = ma to find acceleration.

Okay, so here is what I did:
Total forces of X are Force of friction, force of gravity and force of inertia. I obtained the Normal force by using the the y component of gravity (36 x 9.8(cos10)) and then using that number I obtained the value for the force of friction to be 132.03.
Plugging that into the sum of horizontal forces (as stated above): 132.03 - (mgsin10) -Force i = 0
So the force of i = 70.76 = ma so a = 70.76/36
Is this the correct path?
Again I can't stress how thankful I am for your assistance.
 
  • #8
soul_observer said:
Okay, so here is what I did:
Total forces of X are Force of friction, force of gravity and force of inertia. I obtained the Normal force by using the the y component of gravity (36 x 9.8(cos10)) and then using that number I obtained the value for the force of friction to be 132.03.
Plugging that into the sum of horizontal forces (as stated above): 132.03 - (mgsin10) -Force i = 0
So the force of i = 70.76 = ma so a = 70.76/36
Is this the correct path?
Again I can't stress how thankful I am for your assistance.

That looks right to me.:smile:
 
  • #9
Great! Thanks you guys, you're awsome!
 

Related to Finding maximum acceleration of truck without having package fall off.

1. How do you determine the maximum acceleration a truck can have without having a package fall off?

The maximum acceleration a truck can have without having a package fall off is determined by a combination of factors such as the weight and size of the package, the type and condition of the truck, and the road conditions. It requires careful calculation and testing to find the exact maximum acceleration for each specific scenario.

2. Why is it important to find the maximum acceleration of a truck without having a package fall off?

It is important to find the maximum acceleration of a truck without having a package fall off because it ensures the safety of the package and other vehicles on the road. If the truck accelerates too quickly, the package may shift or fall off, potentially causing damage or accidents. Finding the maximum acceleration also helps to optimize the delivery process and minimize the risk of package loss or damage.

3. What methods are used to find the maximum acceleration of a truck?

There are various methods that can be used to find the maximum acceleration of a truck without having a package fall off. One method is to use mathematical calculations based on the weight and dimensions of the package, the friction between the package and the truck, and the acceleration capabilities of the truck. Another method is to conduct physical tests using different acceleration levels and monitoring the package for any signs of shifting or falling off.

4. How can you ensure the accuracy of the maximum acceleration calculation?

To ensure the accuracy of the maximum acceleration calculation, it is important to have precise measurements of the weight and dimensions of the package, as well as the capabilities of the truck. It is also crucial to conduct multiple tests and factor in different variables such as road conditions and weather. Additionally, using advanced technology and simulation software can also help to improve the accuracy of the calculation.

5. Can the maximum acceleration of a truck change over time?

Yes, the maximum acceleration of a truck can change over time. Factors such as wear and tear of the truck, changes in road conditions, and variations in the weight and size of the packages being transported can all affect the maximum acceleration. It is important to regularly re-evaluate and recalculate the maximum acceleration to ensure the safety of the packages and other vehicles on the road.

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