Newton’s Second Law
1. The equation the rest of the unit is made of
When the arrows do not add to zero, the system accelerates, and the second law says exactly how much.
asys = ΣF / msysand a points the same way as ΣF
Three claims are packed in there, and it is worth separating them. The acceleration is proportional to the net force. It is inversely proportional to the mass. And it points in the direction of the net force, which is not usually the direction of the motion.
That last one does most of the damage. A ball thrown upward is moving up and accelerating down. A car braking is moving forward and accelerating backward. The arrow you compute from the diagram tells you how the velocity is changing, and it is under no obligation to agree with the velocity itself.
A cart on a frictionless floor, photographed every half second, with the graph of acceleration against net force beneath it. Change the mass and watch what happens to the line: the mass is not a point on the graph, it is the slope.
2. Which mass, and which acceleration
The law is written for a system, and both symbols have to refer to the same one. If the boundary holds two blocks, then msys is the total mass and ΣF counts only the forces crossing the boundary. If it holds one block, both change together.
And the acceleration is the acceleration of the center of mass. That is the thread running from Topic 2.1 through all three laws: the parts of a system can do all sorts of things, and the one point that obeys this equation is the center of mass.
3. Two blocks, one string
Here is the arrangement that turns up on every exam. A block on a smooth table is joined by a string over a pulley to a block hanging off the edge. Nothing about it is new: it is Topic 2.1 choosing a system, Topic 2.3 supplying a tension that is the same at both ends of an ideal string, and this topic doing the arithmetic.
Take both blocks as one system and the tension is internal, so the only external force along the direction of motion is the weight of the hanging block, and the mass being accelerated is the mass of both. Then go back to a single block to get the tension.
a = m2g / (m1 + m2)then FT = m1a
Change either mass and watch the last readout, which compares the tension with the weight of the hanging block. It is never one. Ask yourself what would have to be true of the motion for it to be one.
4. Getting a number out of a diagram
The sequence never changes. Choose the system and draw the boundary. Draw one dot and one arrow per force. Tilt the axes so one of them lies along the acceleration. Write the second law on each axis, remembering that the equation across the acceleration reads zero. Solve in symbols. Then put numbers in, and check the direction of the answer against the picture.
The check at the end is worth the ten seconds. An acceleration that comes out negative when you expected the block to speed up usually means a sign was assigned before a direction was chosen.
5. Reading it wrong
Using the weight instead of the net force. The left side of the equation is the sum of every arrow, not the largest one. A block on an incline does not accelerate at g.
Assuming the acceleration points along the motion. It points along the net force. Any object slowing down is a counterexample.
Treating the tension as the weight of the hanging mass. They are equal only when nothing accelerates. If the string were cut the hanging block would fall at g; the string is what stops that, and it can only do so by pulling with less than the full weight while the block is still descending.
Check yourself
1. A 1500 kg car speeds up from rest to 27 m/s in 9 s. Find the net force on it, and name the object that exerts the forward force.
a = 27/9 = 3 m/s², so ΣF = 1500 × 3 = 4500 N forward. The forward force is friction from the road on the tires. The engine is inside the car and cannot accelerate it, which is Topic 2.3 again.
2. A 4 kg block on a frictionless table is joined over a pulley to a 2 kg hanging block. Find the acceleration and the tension, and explain why the tension is not 19.6 N.
Both blocks: a = (2)(9.8)/6 = 3.27 m/s². The block on the table: FT = 4 × 3.27 = 13.1 N. It is not 19.6 N because the hanging block is accelerating downward, so the forces on it do not balance. A tension equal to its weight would leave it in equilibrium and nothing would move.
3. A ball is thrown straight up. Give the direction of its velocity and of its acceleration on the way up, at the top, and on the way down.
Velocity up, then zero, then down. Acceleration down the whole time, with the same size throughout, because the only force is the weight and it never changes. The moment at the top is the useful one: zero velocity with a nonzero acceleration, which is impossible if you believe the two are the same thing.
4. The same net force is applied to a 2 kg cart and a 6 kg cart. Compare their accelerations, and compare the distances they travel in the first 2 s from rest.
The light cart accelerates three times as fast. Distance from rest is ½at² with the same t, so it also travels three times as far. The ratio of the distances is the inverse ratio of the masses, which is a quick check on any two-cart question.
5. A student analyzing two blocks joined by a string writes ΣF = m2g and m = m2, then is surprised to get a = g. Diagnose it in one sentence.
The force came from the two-block system, where the weight of the hanging block is the only external force along the motion, and the mass came from the one-block system. Use the total mass with that force, or use a single block and include the tension, but do not mix them. The answer a = g is the giveaway: it describes a block with nothing attached to it at all.
Topic 2.6, Gravitational Force. The second law needs forces to work on, and the next four topics supply them. First the one that acts across empty space and gives every object on Earth its weight.