I'm working through old past papers for my current maths course. I've gotten to the following question, but nowhere in my notes does it mention acceleration. I've posted a picture of my half attempt but im not sure where to go from there, or if its even right
Thanks for any help
Regards
Mike

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You're using a supposed $F$ (friction, I guess) instead of calculating acceleration. You're actually not answering section (a) concerning the normal reaction. – Joffan Dec 22 '16 at 11:33
1 Answers
Firstly, it's important to note that in almost every case, the use of the word 'smooth' in such questions means produces no friction, so you can ignore the effect of friction, $F$, and simplify your diagram.
Your analysis at the bottom, beginning $-F + 12 \cos(-20) = 0$, suggests that the forces here are balanced, that is the stone will not move at all (but we are told that it does, because you're pushing it!).
The missing component, and perhaps the most important equation in Newtoniam mechanics is known as Newton's Second Law: $$F = ma$$ Where $F$ is the net force applied to a particle, $m$ its mass and $a$ is the resultant acceleration. (Note this equation deals in one dimension, but is applicable in and number of dimensions as $ \textbf{F} = m \textbf{a}$ in vector form).
I am slightly worried that your course, which expects solutions to these problems, does not anywhere mention acceleration - it's really very fundamental! Can I enquire as to what type of course and where? I may be able to point you to better elementary mechanics resources.
edit: Note that my $F$ is not the same as yours!!
For part (a) your expressions are correct, though I think the use of $-20 ^{\circ}$ in the negative doesn't help your understanding. Perhaps a clearer way to resolve this component-wise is to note that the component of the push $P$ in the vertical $j$ direction is $|P|\sin(20)$ and then note this is downwards and thus we get $-|P|\sin(20)j$. We have, for all real $x$ that $\sin(-x) = -\sin(x)$ so the two expressions are indeed the same, but I find thinking it through in this way to be a good sense-check.
Newton's Laws of Motion
Following comments, and related to your question. (Paraphrased from Wikipedia, but many other better explanations are available just by googling.)
First Law: An object will remain at rest, or travelling in a straight line at a constant speed, unless acted upon by a net force.
In your question, for instance, writing $-F + 12 \cos(-20) = 0$ suggests the horizontal forces sum to zero, i.e. there is no net force and hence no movement.
Second Law: The acceleration of a particle acted upon by a net force is proportional to the magnitude of that force, and is in the same direction as that force.
This is your $F = ma$ and is the absolute key to all these problems.
Third Law: A body exerting a force on another body experiences an equal and opposite force on itself.
This is where the reaction force comes from; the stone pushes down on the floor with force $-mg.i$, so the force pushes back up with a Normal Reaction Force of $+mg.i$.
I hope these can help!
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Hi Its a maths course with the ou... It has mentioned acceleration in the past, but its not mentioned anywhere in statics only the acceleration due to gravity – Michael Williams Dec 22 '16 at 11:59
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Hmm, OU usually provides pretty complete courses, I'm surprised that this wasn't a big feature. I would advise looking in to Newton's Laws of Motion, and using them to form the crucial basis of each of your answers. I'll add them to the above answer. – benlavelle Dec 22 '16 at 12:10
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Is the second part of the question as simple as dividing 18 by 12? – Michael Williams Dec 22 '16 at 12:18
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Not quite - the acceleration is the horizontal direction, so you only want the component of the force in the i direction that you worked out before. There is no acceleration in the vertical direction because the Normal Reaction, N, balances the vertical component (by N3L), so there is no net force and thus no movement in this direction (by N2L). – benlavelle Dec 22 '16 at 12:29
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Do you know you said I worked it out before (the horizontal direction). Is that the 4.9N? I thought that was wrong because I put it equal to 0 and its moving not stationary – Michael Williams Dec 22 '16 at 13:15