# Work & Inclided Plane

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

2. Homework Equations
$$W = F_{\parallel}d$$
$$F_{fr} = \mu F_N$$

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

$$F_y = F_g$$

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

and since $$h = d\sin(\theta)$$,

$$W = \frac{F_g h}{{\sin}^2(\theta)} = \frac{mgh}{{\sin}^2(\theta)}$$

I think the answer should be $$mgh$$? What did I do wrong?

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## Answers and Replies

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what about the horizontal component?

Why would I need the horizontal component? I already calculated the force $$F$$ which is being exerted in the direction of motion of the car.

PhanthomJay
Homework Helper
Gold Member
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.

Why is that assumption wrong?

cristo
Staff Emeritus
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

$$F_y = F_g$$

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

and since $$h = d\sin(\theta)$$,

$$W = \frac{F_g h}{{\sin}^2(\theta)} = \frac{mgh}{{\sin}^2(\theta)}$$

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

Dorothy

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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.

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.).

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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