Calculating the Minimum Work Needed to Push a Car Up an Inclined Plane

[IMGOOD]

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 $$\mu$$

Homework Equations

$$W = F_{\parallel}d$$
$$F_{fr} = \mu F_N$$

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?

 

[JJ420]

what about the horizontal component?

 

[IMGOOD]

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]

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.

 

[IMGOOD]

Why is that assumption wrong?

 

[cristo]

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.

 

[Dorothy Weglend]

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

 

[AngeloG]

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.

 

[IMGOOD]

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.).

 

[chrisfnet]

I have an identical problem. I'm not sure where to proceed from where I left off, though. Could anyone help?