# Forces on a simple flat slope [closed]

This diagram is of a block (b) moving down a ramp unaided.

Taking gravity (g) as $9.8ms^{-2}$, and the mass of the block (b) as $0.2kg$, how can I find the force acting along the slope (f).

BTW: the ramp is inclined 1.8 degrees, not radians.

I thought I could use basic trigonometry, and $F=ma$:

$$Sin(1.8) = \frac{0.2*9.8}{f}$$ $$f = \frac{0.2*9.8}{sin(1.8)}$$

However this results in: $f = 62.40$

Seeing as force is a vector I thought I could use trig to split into it's components but obviously I'm doing something wrong.

Sorry for the noob question, but it's got me clueless.

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## closed as off-topic by tpg2114, Qmechanic♦Nov 3 '13 at 2:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

HINT: what happens when you let the angle go to zero? Is the result you get for $f$ physical? – Marek Mar 20 '11 at 22:52
HINT #2: which one of the forces $f$, $g$ is more fundamental and which is derived? What does it tell you about their relative magnitudes? – Marek Mar 20 '11 at 22:56

When in doubt, always draw a force diagram. Remember, $g$ will always point straight down, $f$ will always run parallel to the plan upon which you're moving, and your third vector will always be perpendicular to the plane. With that in mind, your problem will have the following force diagram:

In this case, $a$ is the force component of gravity perpendicular to the plane. If you were working with friction, it would be important ... but you're not, so feel free to ignore it. The angle directly opposite $f$ is $1.8^{\circ}$.

You can use the law of sines (basic trig) to calculate f at this point.

$$\frac{f}{sin 1.8} = \frac{g}{sin 90}$$

Or ...

$$f = g * sin 1.8$$

I'll let you plug in the numbers (gravity and the mass of the block) to solve for $f$.

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@EAMann: we prefer not to have complete solutions given for homework questions. It might be appropriate to edit some of the detail out of your answer (although it is a good answer). – David Z Mar 21 '11 at 15:56
@David. It's not exactly a homework question. The question is from a year 12 paper that te teacher gave me now in year 11. So it's not like I'm being marked it's just to learn how the questions are different. (I get the forces and vectors, but I don't know how exactly to use them). The answer is great, thanks – Jonathan. Mar 21 '11 at 16:01
@Jonathan: ah, OK. For future reference it might be good to mention these things in the question. But FWIW, the policy on homework questions really applies to all "homework-type" questions, basically anything that's being done as an exercise, in which the point is to understand the method of solution rather than just to get the answer. – David Z Mar 21 '11 at 16:03
@David - Can you add that to the site FAQ? Not something that occurred to me when I first came in to the site. And it's not the kind of thing I'd scour meta for before posting an answer. – EAMann Mar 21 '11 at 16:05
@EAMann: that would be nice to have on the FAQ, but I don't have the ability to edit it. Now that I look at it, FYI, it looks like our homework policy is pretty much defined by this question and this one on meta. – David Z Mar 21 '11 at 16:09

Here's something that might be worth remembering: when you split a vector into components (as you were trying to do), the original vector is always the hypotenuse. See if you can use that fact to find the error you made in writing the formula

$$Sin(1.8) = \frac{0.2*9.8}{f}$$

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But here g is opposite the angle? – Jonathan. Mar 21 '11 at 7:39
The gravitational force can't be opposite the angle if it's the hypotenuse. That's the point: you have to draw the triangle in such a way that the original vector (the gravitational force, here) is the hypotenuse. Don't just use whatever triangle happens to exist in the problem. – David Z Mar 21 '11 at 16:05
But what if I was given the force down the ramp (f) and told to work out the gravitational force? Which would be the original? – Jonathan. Mar 21 '11 at 17:25
@Jonathan: the gravitational force would still be the original. You can use the word "total" rather than "original" if it makes more sense to you. (Or even "resultant" if you're familiar with that term) – David Z Mar 21 '11 at 18:39