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I was reading Introduction to Classical Mechanics (Morin) recently and came across this example, which said that the friction force (marked $F_f$) is $Mg\sin\theta - Mg\cos\theta$. However, I can't see why this would be: The leg of the triangle bounded by $Mg$, $F_f$, and the dotted line on the top right hand side seems to show that $F_f = Mg\cos\theta - Mg\sin\theta$, subtracting the force on the other triangle with angle $\theta$ on the bottom left hand side, which seems to be clearly $Mg\sin\theta$.

Where am I wrong?

Edit: I'm sorry for not presenting the problem as shown in Morin. The $Mg$ arrow to the right is the applied force. Here is the exact question:

A block of mass $M$ rests on a plane inclined at angle $\theta$ (see Fig. $1.2$). You apply a horizontal force of $Mg$ to the block, as shown.

Assume that the friction force between the block and plane is large enough to keep the block at rest. What are the normal and friction forces (call them $N$ and $F_f$ ) that the plane exerts on the block?

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  • $\begingroup$ What in the world is that horizontal arrow that points to the right? That's the strangest free body diagram that I've ever seen published. Pretend that arrow is not there; perhaps that will make your analysis more clear. Also, your solution makes no mention of the coefficient of friction. $\endgroup$ – garyp Jul 13 '16 at 11:20
  • $\begingroup$ I read more slowly, and I see why there is no mention of the coefficient of friction. :) But I still don't understand what's going on in this example. $\endgroup$ – garyp Jul 13 '16 at 11:26
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I think the difficulty is that you are not stating the whole of the problem as given by Morin. You are only giving us part of it, out of context. That is why it is confusing.

The puzzling force $Mg$ acting horizontally to the right (which I shall call $P$) is probably the applied force mentioned in the title. Giving $P$ the value $Mg$ seems designed to cause confusion!

(You do not say if $P$ is the minimum force required to keep the block from slipping downhill or the maximum before it starts moving uphill - this information will be needed to work out the coefficient of limiting static friction.)

Assuming the block is in static equilibrium, the net force acting up the plane is :
$$F_f + P\cos\theta - W\sin\theta = 0$$ where $W=Mg$ is the weight of the block, applied force $P=Mg$ also, and $F_f$ is the friction force (assumed to act up the plane; if -ve it acts downhill). This rearranges to give the equation you require.

Your mistake seems to be that you are interpreting $F_f$ as the component of $P$ acting up the incline and writing $F_f = P\cos\theta = Mg\cos\theta$. However, $F_f$ and $P$ are not related in this way.

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  • $\begingroup$ "designed to cause confusion" and how!!. I hope the OP responds and writes the entire question as presented in Morin. $\endgroup$ – garyp Jul 13 '16 at 13:20
  • $\begingroup$ Sorry for the confusion! I just edited the question with the exact wording. sammy-yes, the Mg is the applied force. $\endgroup$ – Shreyas B. Jul 13 '16 at 17:01
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Setting up Newton's 1st law along the slope:

$$\sum F=0\\ F_f-(Mg)_{x, left} +(Mg)_{x, right}=0 \\ F_f-Mg\sin(\theta) +Mg\cos(\theta) =0 \\ F_f=Mg\sin(\theta) - Mg\cos(\theta) $$

I don't see the confusion. Maybe you tried to solve this in your head, which usually makes one stuck in believing the signs are wrong along the way because some final rearranging is missing. Maybe the terms were not moved to the other side of the equal sign? Always write out the full equation.

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  • $\begingroup$ What do your subscripts "left" and "right" mean? $\endgroup$ – garyp Jul 13 '16 at 11:23
  • $\begingroup$ @garyp There are two $Mg$-forces in the figure. $\endgroup$ – Steeven Jul 13 '16 at 12:42
  • $\begingroup$ Yes, and I'm confused about that. $Mg$ sounds like it's weight, but what weight points horizontally? If it's a different force, why label it $Mg$? I don't understand the diagram or the problem. $\endgroup$ – garyp Jul 13 '16 at 13:17
  • $\begingroup$ @garyp Maybe the problem states that there is some applied force equal to gravity? It could be any setup. But what does it matter? $\endgroup$ – Steeven Jul 13 '16 at 13:22
  • $\begingroup$ You are probably right. The only reason in matters to me is that I had no conception that the right-pointing arrow was another force. I thought it was some kind of graphical aid to doing the trigonometry ... rotating the force of gravity for some reason. The original wording would have been very helpful. $\endgroup$ – garyp Jul 13 '16 at 13:42

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