# Mechanics Question

A particle $P$ of mass $0.6\text{ kg}$ moves upwards along a line of greatest slope of a plane inclined at $18°$ to the horizontal. The deceleration of $p$ is 4 ms−2

The first part asks us to find the frictional and normal components of the force excreted on $P$ by the plane. Then they ask us to find the coefficient of friction between $P$ and the plane.

Now I understand that this has to be done by $f_{net} = ma$, but the solution at the back only uses frictional force and the component of the weight down the slope. I don't understand why there isn't a force acting up the slope. Say this force is $K\text{ N}$, then shouldn't the $f_{net}$ equation be $$K- mg \sin18- \text{frictional force}= 0.6(-4)$$

• Did you mean the decelaration is $4\text{ m.s}^{-2}$? Can you draw the force diagram? Apr 10 '14 at 19:23
• the decelaration is 4 m.s−2 Apr 10 '14 at 19:24

I think you are mixing up speed and acceleration. The ball is moving up and thats why friction should point down the slope, but if there were a force acting up the slope then it would mean that $P$ is actively using energy to climb up. For large enough $K$, $P$ would be accelerating upwards, as if you were pushing it.

I believe the situation is instead more like you gave the ball an initial velocity and this is the kinematics of the ball experiencing only friction (and the effect of its weight) along the slope, causing a deceleration. Hopefully this convinces you that $K=0$, and $f_{net} = ma$ means

$$-0.6 g \sin 18º - \text{frictional force} = 0.6 \times (-4)$$

• No problem :) Hey I just noticed that you didn't accept the answers to the questions you asked. If they answered the question, you should do that by clicking on the tick below the voting buttons for that answer. Apr 10 '14 at 21:28
• I didn't know about that, sorry! Apr 11 '14 at 6:54
• np, the rules or something I read on this site said that if you don't accept then some people might stop answering you so I just thought I should tell you. Apr 11 '14 at 8:24
• @CalvinKhor I think you gave the wrong solution. Let's say $K=0$ and $mg\sin18^\circ>f_{sm}$ then the particle will move down. Hence, the Newton's second law will be $$mg\sin18^\circ- f_k=ma.$$ Apr 11 '14 at 15:35
• I don't think so; the force does not determine the speed, only the acceleration. It is supposed to decelerate, so it works out. If I'm still wrong, would you care to elaborate? Apr 11 '14 at 16:36

Using Newton's second law of motion: \begin{align} \sum F&=ma\\ K-mg\sin18^\circ\pm f_k&=ma\\ K-mg\sin18^\circ\pm\mu_k N&=ma\\ K-mg\sin18^\circ\pm\mu_k mg\cos18^\circ&=ma.\\ \end{align} If $0\le K<mg\sin18^\circ+f_{sm}$, the sign of the friction $f_k$ should be positive, why? If $K>mg\sin18^\circ+f_{sm}$, the sign of the friction $f_k$ should be negative, why? What would happen to particle if $K=mg\sin18^\circ$? Hint: Draw the force diagram and determine the particle's movement from the 3 cases.

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Where would a force acting up the slope come from? You resolve gravity into the force down the slope and force into the slope. The force into the slope creates a frictional force that is also down the slope. Your $K$ is zero, then your equation is correct.