# Finding maximum acceleration of truck without having package fall off.

#### soul_observer

1. 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?

2. Homework Equations
Force of friction<(=) mu of static friction(normal force)

3. 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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#### PhanthomJay

Homework Helper
Gold Member
1. 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?

2. Homework Equations
Force of friction<(=) mu of static friction(normal force)

3. 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?

#### Stovebolt

1. 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?

2. Homework Equations
Force of friction<(=) mu of static friction(normal force)

3. 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.

#### soul_observer

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

#### soul_observer

So is the Force of inertia the force you are solving in this problem? Would that be the acceleration? I am a bit confused.

#### Stovebolt

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.

#### soul_observer

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.

#### Stovebolt

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.

#### soul_observer

Great! Thanks you guys, you're awsome!

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