Newton's Third Law – Action and Reaction: notes and previous year questions

Answer: Option A

The branch is holding up everything below it: the $200\ \mathrm{N}$ body and the chain's own weight $10 \times 10 = 100\ \mathrm{N}$, so the pull at the top is $300\ \mathrm{N}$. Had the chain been light, the answer would have been $200\ \mathrm{N}$ — the option waiting for anyone who treats every rope as massless out of habit. And a closing word for the whole section: the branch pulls the chain up with $300\ \mathrm{N}$ and the chain pulls the branch down with $300\ \mathrm{N}$, a genuine pair on two different bodies. What holds the chain still is not that pair; it is the first law applied to the chain alone, the branch's pull up against the weight hanging below. A third-law pair is never the reason anything is in equilibrium.

Answer: Option C

$F_{2}$ and $F_{3}$ are perpendicular, so their resultant has size $\sqrt{8^{2}+6^{2}} = 10\ \mathrm{N}$ — and that is $F_{1}$, exactly as it must be, since the particle is at rest and the three arrows close up. Remove $F_{1}$ and that $10\ \mathrm{N}$ resultant is left on its own: $a = \dfrac{10}{5} = 2\ \mathrm{m/s^{2}}$, in the direction $F_{1}$ used to point away from. The $4.8$ on offer is $\dfrac{F_{2}F_{3}}{mF_{1}}$, a combination with no meaning here. Perpendicular forces only ever combine through the square root, never by adding or subtracting their sizes.

Answer: Option C

Choose the block, and it has exactly two arrows: its weight $Mg$ down and the floor's normal reaction $N$ up. The quarter-weight the question quotes is the force the block puts on the floor, which belongs on the floor's diagram — but its third-law partner is the floor pushing back on the block with the same size, so $N = \dfrac{Mg}{4}$. Take down as positive, since that is where the acceleration points: $Mg - \dfrac{Mg}{4} = Ma$, giving $a = \dfrac{3g}{4}$. A normal reaction of a quarter of the weight is no contradiction — $N = mg$ only ever held for a body with no vertical acceleration.

Answer: Option B

One body of mass $m_{1}+m_{2}$, one external horizontal force $F$, so $a = \dfrac{F}{m_{1}+m_{2}}$. The push of each block on the other is in there too, twice, once each way, and the pair sums to zero — the only time third-law partners may share a diagram, because that diagram is of both bodies at once. What this view cannot tell you is how hard the two blocks press on each other. For that they have to be cut apart.

Answer: 550 N

The vertical part of the push is $F\sin 30^{\circ} = 100 \times 0.5 = 50\ \mathrm{N}$, pressing down, and the roller is not accelerating vertically, so $N = mg + F\sin\theta = 500 + 50 = 550\ \mathrm{N}$. The ground is carrying $50\ \mathrm{N}$ more than the roller weighs, and that extra is precisely what the hands are adding. Only $F\cos 30^{\circ} \approx 86.6\ \mathrm{N}$ of the push is doing anything to move it forward; the rest is spent pressing it down.