Forces on Accelerating Objects

Add the forces first; the leftover is what accelerates the object

An accelerating object is one whose forces don't cancel, there is a leftover, the net force, and the object accelerates in that exact direction.

To analyse any accelerating object, work in two steps. First, draw a free-body diagram and add every force as a vector to get the net force $F_{net}$. Second, the object accelerates in the direction of $F_{net}$, with size set by $F_{net} = ma$. No single force, not the engine's push, not gravity alone, sets the acceleration; only the resultant does.

• Forces balanced ($F_{net} = 0$): no acceleration, at rest or constant velocity.
• Forces unbalanced ($F_{net} \neq 0$): acceleration in the direction of $F_{net}$.

Only the net force accelerates an object. Add all forces as vectors to get $F_{net}$; the object accelerates in the direction of $F_{net}$ with magnitude given by $F_{net} = ma$. If $F_{net} = 0$, the object stays at rest or moves at constant velocity.

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From a free-body diagram to a number for the acceleration

We just saw that the net force sets the acceleration. That raises a question: how do we turn the forces on a real object into an actual value for $a$? This card answers it → resolve the forces along the line of motion, find $F_{net}$, then apply $F_{net} = ma$.

Once you have the net force along the direction of motion, the second law is just one rearrangement away from the answer you want.

Newton's second law, from the NESA formula sheet, is:

Rearrange for whichever quantity is unknown: $a = \dfrac{F_{net}}{m}$ or $m = \dfrac{F_{net}}{a}$. Force is in newtons (N), mass in kilograms (kg) and acceleration in $\text{m s}^{-2}$.

To find acceleration: add forces along the line of motion to get $F_{net}$, then use $a = \dfrac{F_{net}}{m}$. The acceleration points in the direction of the net force. For the example, $F_{net} = 20 - 8 = 12\ \text{N}$, so $a = 12/4 = 3.0\ \text{m s}^{-2}$.

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Why a rising lift makes the scales read more than your weight

We just learned to compute acceleration from the net force. That raises a question: what does the second law tell us when the motion is vertical, like standing in an accelerating lift? This card answers it → the floor's normal force must do more than balance weight; the extra produces the net upward force.

When a lift accelerates upward, the floor has to push you up harder than gravity pulls you down, and that larger push is exactly what the scales read.

Stand on scales in a lift. The forces on you are your weight $W = mg$ downward and the normal force $N$ from the floor upward; the scales read $N$. When the lift accelerates upward with acceleration $a$, the net upward force is $N - mg$, so by the second law:

Because $N = m(g+a) > mg$, the scales read more than your true weight while the lift accelerates upward. When the lift accelerates downward, $N = m(g - a) < mg$ and you feel lighter. Your mass never changes, only the net force, and therefore the normal force, does.

In a lift accelerating upward, the scale reading (normal force) is $N = m(g+a)$, larger than your weight $mg$; accelerating downward gives $N = m(g-a)$, smaller. Mass is unchanged, the net force changes, which changes the support force the floor must provide.

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