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Work & Inclided Plane

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1. Homework Statement
What is the minimum work needed to push a car of mass [tex]m[/tex] kilograms, an absolute distance of [tex]d[/tex] meters (or a height of [tex]h[/tex] meters) up a [tex]\theta[/tex] degree incline plane. a) Ignore Friction. b) Assume that the effective coefficient of friction is [tex]\mu[/tex]

2. Homework Equations
[tex]W = F_{\parallel}d[/tex]
[tex]F_{fr} = \mu F_N[/tex]

3. The Attempt at a Solution
For part a), if you want to push the car up the inclined plane, then you would need to apply a vertical force with a magnitude equal to that of gravity, right? So you would want

[tex]F_y = F_g[/tex]

[tex]\Rightarrow F\sin(\theta) = F_g \Rightarrow F = \frac{F_g}{\sin(\theta)}[/tex]
where [tex]F[/tex] is a force in the direction of motion of the car. So the work needed to push the car would be
[tex] W = Fd = \frac{F_g}{\sin(\theta)} d[/tex]

and since [tex]h = d\sin(\theta)[/tex],

[tex] W = \frac{F_g h}{{\sin}^2(\theta)} = \frac{mgh}{{\sin}^2(\theta)}[/tex]

I think the answer should be [tex]mgh[/tex]? What did I do wrong?
 
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Answers and Replies

20
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what about the horizontal component?
 
32
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Why would I need the horizontal component? I already calculated the force [tex]F[/tex] which is being exerted in the direction of motion of the car.
 
PhanthomJay
Science Advisor
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You have mistakenly assumed that F_y = F_g. Rather than over complicate the problem, try using work -energy methods and note the the minimum work required with or without friction occurs when the block is moved at constant velocity.
 
32
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Why is that assumption wrong?
 
cristo
Staff Emeritus
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First draw a free body diagram. In order to push the box up the slope, apply a force F parallel to the slope. Then resolve the weight into parallel and perpendicular components. You will find a familiar answer will drop out easily using this method.
 
3. The Attempt at a Solution
For part a), if you want to push the car up the inclined plane, then you would need to apply a vertical force with a magnitude equal to that of gravity, right? So you would want

[tex]F_y = F_g[/tex]

[tex]\Rightarrow F\sin(\theta) = F_g \Rightarrow F = \frac{F_g}{\sin(\theta)}[/tex]
where [tex]F[/tex] is a force in the direction of motion of the car. So the work needed to push the car would be
[tex] W = Fd = \frac{F_g}{\sin(\theta)} d[/tex]

and since [tex]h = d\sin(\theta)[/tex],

[tex] W = \frac{F_g h}{{\sin}^2(\theta)} = \frac{mgh}{{\sin}^2(\theta)}[/tex]

I think the answer should be [tex]mgh[/tex]? What did I do wrong?
Assuming you are pushing parallel to the incline, the force is just [itex] F [/itex], the force of gravity that you are pushing against is the component of gravity along the incline, or [itex] F_g(\sin \theta)[/itex].

Dorothy
 
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103
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You're pushing a car up an inclinced plane =).

Remember, your normal force is perpendicular to the object that is at rest. So your normal force isn't at a 90 degree angle =). So think about that.

Also, as said draw a free body force diagram. It will help get you with how the forces are working and balancing out in the entire picture.
 
32
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Thank you very much you guys. Thanks to you guys, I figured out what was wrong with my analysis.

First of all, my choice of the coordinate system wasn't a very smart one -- I chose the y-axis to point directly upwards and the x-axis to point directly to the right.

Second of all, I was thinking that there was only one vertical force that was acting on the car -- gravity. But I forgot to take into account the vertical component of the normal force (I hadn't taken into account this vertical component previously because I thought it canceled out with one of the components of gravity perpendicular to the inclined plane. But in my coordinate system there was no such component! -- there was only one component of gravity which was acting in the -y direction.).
 
39
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I have an identical problem. I'm not sure where to proceed from where I left off, though. Could anyone help?
 

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