-6
$\begingroup$

Please note that it is stating force A is equal to the component of another force, not magnitude. Let me say this again. You are not going to say component is the magnitude.

However, even when I asked the department chair of physics, he said it is asking about magnitudes, and that I should know it is saying magnitude.

I understand the answer (d) is wrong, you don't need to explain it. If it is really my professor's fault that answer c is wrong. My grade can be went up one level :D However, my professor is very angry when I ask this question. I understand the answer never must true, but it doesn't mean it always false. Moreover; Answer c is always false.

The problem is:

Which of the following statements is true about an object at rest on an inclined plane?

(a) There is no friction

(b) The friction force is equal to the object's weight

(c) (correct) The friction force is equal to the component of the weight pointing down the ramp

(d) (my original choice) The friction force equal to $\mu_s N$

In explaining why (c) is correct, my professor said, "if two forces are not equal, how can the object is at rest?" He didn't mention the word "magnitude" at all. One again, answer C is talking about the component of another force, not magnitude.

Professor told me if the object is at rest, two forces are same. He didn't use the word "magnitude" at all. I guess my professor thinks different between two forces that the forces exerts on a rest object is only the negative sign.

I am just a little student in ELAC of a Physics 20 class at the summer.


enter image description here Updated: My professor's explanation: what my professor told me about why answer C is totally correct, ''the word 'components' can be rounded to be magnitude as same as 3.06 rounds to 3.1. This means components is magnitude''. I believe my professor.'

$\endgroup$

closed as unclear what you're asking by DanielSank, Kyle Kanos, John Rennie, peterh, WillO Aug 8 '17 at 7:40

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Can you explain your second bullet (v2)? You seem to prove that the friction force must be equal to the component of the weight pointing down the ramp, but somehow you draw the opposite conclusion. Everything you stated supports the idea that they must be equal. $\endgroup$ – Cort Ammon Aug 7 '17 at 22:38
  • 3
    $\begingroup$ It seems to me like this whole thing is about whether the word "magnitude" should be there somewhere. The only answer I can give is that yes, technically it should, but using technically correct language all the time gets boring. $\endgroup$ – Javier Aug 7 '17 at 22:41
  • $\begingroup$ Get used to not all exam questions being perfectly phrased and get used to not all "correct" answers being actually correct. Sometimes, to pass, you need to know what answer is expected, which may not be the most correct answer. Education is not about learning things. It is about learning how to learn. Because, if you really want to learn, you'd be learning your whole life. $\endgroup$ – safesphere Aug 7 '17 at 23:58
  • 1
    $\begingroup$ Technically, you are correct, but if there must be a correct choice among the 4 possibilities, (c) is the one to choose. I agree, two vectors are equal if and only if they have equal magnitudes and point in the same direction. If an object is in static equilibrium, the sum of forces must be zero. Two equal vectors cannot add to zero. $\endgroup$ – Bill N Aug 7 '17 at 23:59
  • 1
    $\begingroup$ If you thought (d) was correct, then you were already thinking about magnitudes. After all, you surely didn't think the frictional force pointed in the same direction as N! $\endgroup$ – Jahan Claes Aug 8 '17 at 0:51
-1
$\begingroup$

It sounds like your professor is resolving the forces in axes parallel to the ramp. In this case the component of the weight in the (down) ramp direction is exactly equal to the friction. If you are talking about components, you do not have to use the word "magnitude".

$\endgroup$
  • $\begingroup$ I disagree. A component is also a vector. $\endgroup$ – Bill N Aug 8 '17 at 0:06
  • $\begingroup$ It's opposite to the friction. The weight is in the down-ramp direction; the friction is in the up-ramp direction. Even if you're talking about components, the two forces differ by a sign, and you have to take the magnitude. $\endgroup$ – Jahan Claes Aug 8 '17 at 0:50
0
$\begingroup$

The problem with choice (d) is the word equal. The friction force will in all likelihood be less than $\mu_s N$, rarely if ever will it be equal to $\mu_s N$.

$\endgroup$
  • $\begingroup$ Did I say I know answer d is incorrect? $\endgroup$ – Jonathan Sum Aug 8 '17 at 3:21
  • $\begingroup$ Yes, but you didn't say that you know why it's wrong. I guess I don't understand what your question is. $\endgroup$ – garyp Aug 8 '17 at 11:40
  • $\begingroup$ My question is simple. Answer C is wrong too. $\endgroup$ – Jonathan Sum Aug 8 '17 at 13:41
  • 1
    $\begingroup$ I can understand your frustration. Yes, all the answers are wrong, and you have every reason to be upset by that. I will say that the way you worded your question made it hard to understand what you were asking. It's hard to write a clear, unambiguous question. Note that the reason your question was closed was because it was unclear what you were asking. Don't be pissed off at the people here trying to help. $\endgroup$ – garyp Aug 8 '17 at 13:42
  • $\begingroup$ Again, your phrasing is not clear. (c) is wrong because the force of friction is up the ramp, but the force of gravity is down the ramp. The word "component" can mean "vector component along a coordinate axis" or it can mean "the scalar that multiplies a coordinate unit vector". One can't depend on a standard definition for the term, so the question is ambiguous. However regardless of which definition you choose, they are not equal in this problem. Can you assume the word "magnitude" is implied? Not really. But multiple choice is often about the best, not the right answer. $\endgroup$ – garyp Aug 8 '17 at 14:55
0
$\begingroup$

The wording of the "correct" response is not an answer to the question although if one has to answer the question one could reason that responses (a), (b) and (d) are definitely incorrect and response (c), although technically incorrect, is all that is left.

A component of a vector is a vector.

Let $\hat i$ be a unit vector pointing down the slope.

In this case the component of the weight down the slope is a vector pointing down the slope.
$W_{\rm down} \, \hat i$ where $W_{\rm down} $ is positive.

The frictional force is a vector.

In this case the frictional force is a vector pointing up the slope.
$F (-\hat i)$ or $-F \hat i$ where $F$ is positive.

Response (c) states that these two vectors are equal.

$W_{\rm down} \, \hat i = -F \hat i \Rightarrow W_{\rm down} = -F$ which cannot be true as both $W_{\rm down} $ and $F$ are positive.

As the OP stated using the word magnitude would have produced a correct response but it would have been verbose.
"(c) The magnitude of the friction force is equal to the magnitude of the weight pointing down the ramp".

All this shows is how difficult it is to write unambiguous and technically correct questions and the need for one to enter into the "spirit" of the question.
After all the OP did not query the use of the undefined $\mu_{\rm s} N$ in response (d).

$\endgroup$
  • $\begingroup$ @JonathanSum If I had to answer the question I would give the response (C) because I would be answering the question in the "spirit" of what I thought the setter of the question was asking - ie asking about the (magnitude of a) force up a slope and the (magnitude of a) force down the slope. As a pedant I would say that wording of the question leaves a lot to interpretation. Passing exams is not just about knowing the Physics but is much about being able to interpret what the examiner intended which is part of so called "examination technique". $\endgroup$ – Farcher Aug 18 '17 at 6:51
-1
$\begingroup$

So, assuming I understand your question (which I may not, I seem to have gotten a bit mixed up in your clarification of what the exact problem with the explanation was), but to the best of my ability, here is the issue. Forget for a second where the words "magnitude" and "component" were used in your conversation with the teacher or from within the problem. For the sake of this answer: magnitude is simply referring to the absolute value off any measurement, regardless of direction, and components are referring to the components of any given force. If I refer to the absolute value of a single component, I will do so by clarifying it as "the magnitude of the component". With that out of the way, here it goes:

When an object (I assume a square or other noncircular object) is sitting on an inclined plane (frame of reference being strictly two dimensional) without moving, there are (in simplified terms) three forces acting upon it. The first is the gravitational force, which is acting directly downwards (regardless of angle of the plane). The gravitational forces magnitude is equal to the mass of the object multiplied by the acceleration (so g = 9.8 m/s/s). The second force acting on the object is the normal force. The normal force is not acting directly upwards, it instead acts in a direction perpendicular to the inclined plane. The third force acting on the object is static frictional force, which is acting in the opposite direction of the possible movement, so in other words, up the ramp.

The gravitational force is split into two components that are relevant to the current situation: one component parallel to the inclined plane (and therefore parallel to the path of possible movement of the object) and the other component is perpendicular to the inclined plane (and acting in the opposite direction the normal force). I will denote these two component forces as Gp (for gravitational component parallel to the plane) and as Gq (for gravitational component perpendicular to the plane). Having defined our variables, here is the situation:

Force total = F(normal) + F(gravitational) + F(frictional)

We then split this equation into two separate considerations: one dimensional movement equation parallel to the inclined plane , and one perpendicular to the inclined plane.

Force parallel = Gp + F(frictional)

Force perpendicular = Gq + F(normal)

The component of gravity acting perpendicular to the inclined plane (Gq) cancels with the normal force. We know that the object does not rise up away from the inclined plane, nor does it sink into it. Therefore we know that, purely in terms of magnitude (absolute value), Gq = F(normal). So the perpendicular aspect of the force equations has been cancelled and is not relevant to the possible movement or stillness of the object. The only equation to concern yourself with is the parallel equation.

Force parallel = Gp + F(frictional)

This is the equation that determines the motion parallel to the plane (whether it slides up or down). Frictional forces are defined as μN, where N is normal force. This equation though, is speaking only about the absolute value, or magnitude, of the frictional force. The direction of frictional force is always opposite the direction of the possible movement. So if the parallel component of gravity (Gp) is trying to pull the object down the ramp, the frictional force acts up the ramp.

Assuming the object is not moving, and that up is positive and down is negative, we can rewrite a few of our equations....(A) = angle of inclined plane:

Force parallel = Gp + F(frictional) = -mgsin(A) + μN = 0

Force perpendicular = Gq + F(normal) = -mgcos(A) + N = 0

The two possible answers you got confused on, (c) and (d), seem to be both true. Each of the two answers are technically true, the issue is more of one with words. And I would have said this at the beginning, if it weren't for the fact that, in your explanation of the situation, it seems you have accidentally gotten confused. The frictional force's magnitude is equivalent to both the magnitude of the Gp (parallel component of gravitational force) and to the magnitude of μN. Of course, that statement is not taking vector directions into account, hence my use of the word "magnitude". Technically speaking, if direction were being included as part of the problem (without them mentioning it), then both answers would be wrong since μN could be in the same direction as N, and the direction down the ramp is obviously incorrect. The question, therefore, was referring only to the absolute numerical value, or magnitude, of the frictional force. If the problem had been written better, and the word "magnitude" had been included somewhere, it definitely would have made this a less tricky (and possibly misleading) question easier to answer. As far as I can tell, it was the wording of the question that tripped you up, and not the actual concepts behind it. Everything above is just to double and triple check your logic so you can be sure (regardless of the question) that you understand the problem.

Your teacher's explanation is true though. Whether or not it is a complete explanation aside, if the angle of the ramp is increased, and the μ stays the same, eventually the object will slip down the ramp. The math works regardless of what way you come at this (so it doesn't matter in terms of the difference between answers (c) and (d)). I hope this was helpful, my apologies for the poor formatting.

$\endgroup$
  • $\begingroup$ Ok, like I said. I understand the answer D is incorrect. I didn't say my teacher's explanation is incorrect. $\endgroup$ – Jonathan Sum Aug 8 '17 at 3:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.