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A farm tractor tows a $3900 kg$ trailer up a $16^\circ$ incline with a steady speed of $3.0 m/s$. What force does the tractor exert on the trailer? (Ignore friction.)

I am having trouble understanding the steps to this problem even thought I know what they are.

They are

1) Calculate the force of the weight pushing down
$3900 * 9.8$ = $38220 $
I get this part

Now here is what I don't get.

2) The force going down :- $38220$ multiplied by the sin of the angle i.e. $38220 \sin 16 = 11k$

What is going through my mind is that the vertical is $38220$ down so by $\sin\theta = \frac{opp}{hyp}$ shouldn't you calculate it by

$h$(being the force) = $\sin 16 (\sin\theta)/38220$(the opposite being the force of gravity)

I'm imagining this triangle upsidown and flipped -- http://revisionworld.co.uk/sites/default/files/imce/trig.gif

the opposite being gravity, the hypotenuse being the force

Thanks!

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I think you may be using the wrong triangle in your approach to this problem.

To find the force that the tractor is exerting on the trailer, you need to find the part of the gravitational force that is pulling the trailer back down the ramp.

To do this, you need to break the gravitational force vector in to components. You want to find the force that gravity exerts parallel to the ramp. So, break the gravity force vector in to components that are parallel or perpendicular to the ramp.

The hypotenuse of the relevant triangle should be the total gravitational force, the adjacent side should be the component of the gravitational force that is perpendicular to the ramp, and the opposite side should be the component of the gravitational force that is parallel to the ramp.

(Also, you correctly wrote the definition of the sine function, however you then proceed to write a statement that would imply: hypotenuse = sin(theta)/opposite, which is incorrect).

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  • $\begingroup$ Thanks! You are totally correct. I realized I was visualizing the wrong triangle after a while of thinking on it. $\endgroup$ – Zachooz Oct 6 '13 at 17:25
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Let me suggest a different way of looking at the problem.

In one second the trailer moves a distance 3m along the slope, so it moves a distance $h$ upwards, where $h$ is simply given by:

$$ h = 3 \sin 16 $$

This means the potential energy of the trailer has increased by an amount $mgh$, where $m$ is the trailer mass and $g$ is the acceleration due to gravity. This increase in energy can only have come from the work done on the trailer by the tractor, and we know that work = force $\times$ distance. Since you know the distance the trailer has moved (3m) you can calculate the force exerted on it.

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