What Are the Forces Acting on Boxes on Inclined Planes?

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Homework Help Overview

The discussion revolves around forces acting on boxes on inclined planes, focusing on various scenarios involving friction, acceleration, and forces. Participants are exploring concepts related to Newton's laws of motion and the decomposition of forces on inclined surfaces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply Newton's second law to different problems involving boxes on ramps, questioning how to decompose forces into tangential and normal components. Some express frustration with understanding the application of equations and concepts like friction and acceleration.

Discussion Status

Some participants have shared partial attempts at solving specific problems, while others are seeking clarification on the application of principles and equations. There is an ongoing exploration of different approaches to the problems, with no clear consensus on solutions yet.

Contextual Notes

Participants mention challenges with understanding the application of equations and concepts, particularly in relation to their teacher's explanations. There is a focus on specific numerical problems, with some participants expressing confusion about the calculations involved.

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3. A 15 kg box slides down a frictionless ramp of 25 degrees. What is its acceleration?
m/s/s

What is the normal force of a 35 kg box sitting on a slope of 14 degrees? (assume the box is not moving).
N

A 45 kg box is sliding at a constant speed down a ramp of 28 degrees.

What is the frictional force on the box?

What is the coefficient of sliding friction?

After an initial push, a box goes up a frictionless incline of 20 degrees.

What is its rate of deceleration?
m/s/s

A 75 kg skier starts at the top of a 15 m incline at an angle of 25 degrees.
Assuming the coefficient of friction is .15, what is his acceleration?).

m/s/s

What is his speed at the bottom of the hill?
m/s

can someone do those for me just so i can figure out the logic like with the equations id ont get it at all thanks
 
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The principle used is called Newton's 2.law of motion, which gives rise to an energy balance sheet often more convenient to work with.

As for how to do this:
Start by decomposing your forces in the tangential and normal directions.
 
i don't know how to do it that why I am asking my teacher is the worst teacher ever she's a nerd who is smart as s*** but can teach for her life and I am really frustrated right now I've been told to use Newtons 2nd law like 20 times now and i don't know how to use it i hae the equations i just don't know how t apply it so i can get the answer
 
Okay, suppose your incline has a positive angle [itex]\theta[/tex] to the horizontal.<br /> That means that the unit tangent vector [itex]\vec{t}[/itex] up along the incline must be [itex]\vec{t}=\cos\theta\vec{i}+\sin\theta\vec{j}[/itex] where [itex]\vec{i},\vec{j}[/itex] are unit vectors in the horizontal and vertical directions, respectively. Agreed so far?<br /> <br /> Furthermore, the unit normal [itex]\vec{n}[/itex] with positive vertical component must therefore be [itex]\vec{n}=-\sin\theta\vec{i}+\cos\theta\vec{j}[/itex]<br /> Agreed?<br /> <br /> But now, we can, if we wish, write [itex]\vec{i}=\cos\theta\vec{t}-\sin\theta\vec{n}, \vec{j}=\sin\theta\vec{t}+\cos\theta\vec{n}[/itex]<br /> Verify this!<br /> <br /> This is basically what you need to proceed..[/itex]
 
For example, the force of gravity, given by [itex]-mg\vec{j}[/itex]
may now be rewritten as [tex]-mg\vec{j}=-mg(\sin\theta\vec{t}+\cos\theta\vec{n})[/tex]
 
i got 1 and 2 but the rest i keep getting wrong


i don't know hwo to get Fk abd Uk
 
Post your ideas for one of the problems, say number 3.
 
i don't know i have

A 45 kg ,a ramp of 28 degrees.

What is the frictional force on the box? ...

coeficient of friction?...

i tried to use ma =mgsin0 - ukFn = mgsin0-ukmgcos0

and got 389.37 for Fk

and i need fk to get coeeficient so i stoped
 

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