# 3.2.1 Constant Mass (F=ma)

N2L states that $\displaystyle {{F}_{{net}}}=\frac{{dp}}{{dt}}$

For a body with a fixed mass m, N2L can be simplified to become $\displaystyle {{F}_{{net}}}=\frac{{dp}}{{dt}}=\frac{{d(mv)}}{{dt}}=m\frac{{dv}}{{dt}}=ma$ ${{F}_{{net}}}=ma$ is the most familiar form of N2L. Everyone knows what to plug into the formula when there is only one mass and only one force in the scenario. The challenge comes when the scenario involves multiple bodies experiencing multiple forces in multiple directions.

Elevator Man Example

A 60-kg man takes the elevator to a higher floor. Calculate the “weight” of the man when the elevator is

1. speeding up (at a constant rate) from rest to 2.0 m s-1 in 4.0 s,
2. moving upward at a constant speed,
3. slowing down from 2.0 m s-1 to rest in 4.0 s.

Solution

There are only 2 forces acting on the man. (1) the downward gravitational pull mg by the Earth and (2) the upward normal contact force N by the elevator floor. While mg = (60)(9.81) = 589 N is a constant force, N is not. Why does the floor bother to push the man upward? Because it wants to keep the man away! And depending on the motion, the amount of N required to keep the man away can be larger, smaller or equal to mg.

a)

Speeding up from rest to 2.0 m s-1 in 4.0 s, the elevator is accelerating at $\displaystyle \frac{{2.0-0.0}}{{4.0}}=0.50\text{ m }{{\text{s}}^{{-2}}}$. For the man to keep up with the elevator, he must be experiencing an upward net force. This means that N must be stronger than mg.

Forming the N2L equation for the man (taking upward as positive), \begin{aligned}({{F}_{{net}}}&=ma)\\N-mg&=ma\\N&=m(g+a)\\&=(60)(9.81+0.50)\\&=619\text{ N}\end{aligned}

With N larger than the usual of 589 N, the man feels heavier than usual. In fact, if the man were standing on a weighing scale in the elevator, there would be a 619 N normal contact force between him and the scale. And the scale would be reading 619 ÷ 9.81 = 63 kg (instead of the usual 60 kg).

b)

For the man (and the elevator) to be moving at constant speed, the net force acting on the man must be zero.

Forming the N2L equation for the man, \begin{aligned}({{F}_{{net}}}&=ma)\\N-mg&=0\\N&=mg\\&=(60)(9.81)\\&=589\text{ N}\end{aligned}

589 N is the normal contact force a 60-kg man is accustomed to. So he feels a “weight” of 60-kg during this part of the elevator ride.

c)

Slowing down from 2.0 m s-1 to rest in 4.0 s corresponds to an acceleration of $\displaystyle \frac{{0.0-2.0}}{{4.0}}=-0.5\text{ m }{{\text{s}}^{{-2}}}$. To have a downward acceleration, Fnet must be downward. This means that mg must be stronger than N.

Forming the N2L equation for the man (taking downward to be positive), \begin{aligned}({{F}_{{net}}}&=ma)\\mg-N&=ma\\N&=m(g-a)\\&=(60)(9.81-0.50)\\&=\text{ }559\text{ N}\end{aligned}

With N smaller than the usual 589 N, the man feels lighter than usual. In fact, if the man were standing on a weighing scale in the elevator, the scale will be reading 559 ÷ 9.81 = 57 kg (instead of the usual 60 kg).

Horizontal Stack Example
Block A, B and C of mass 4.0 kg, 6.0 kg and 2.0 kg respectively (pressing against each other) is pushed along a smooth horizontal surface by an external force of 24 N.

Calculate

a) the contact force between block A and B

b) the contact force between block B and C

Solution

Using Fxy to denote the contact force that block X exerts on Y, we draw the free body diagram for the given scenario.

Notice that we have identified five forces in the picture: 24 N, FBA (force by B on A), FAB  (force by A on B), FCB  (force by C on B) and FCB (force by C on B). Note that FBA and FAB form an action-reaction pair. So do FCB and FCB.

So, which of these forces must be included in your N2L equation? Well, it depends which mass you have chosen to apply the N2L on.

For example, if you are considering blocks A+B+C as a combined 12 kg mass, then 24 N is the only force acting on this 12 kg mass. All the other forces are internal forces. Internal forces always come in equal but opposite pairs, so they always cancel each other out. Internal forces should therefore be left out of the Fnet summation.

On the other hand, if you are considering block C individually as a 2 kg mass, then FBC is the only force acting on this 2 kg mass. All the other forces do not act on this 2 kg mass and must be left out of the Fnet summation.

a)

If we consider all three blocks as a combined 12 kg mass, then \begin{aligned}({{F}_{{net}}}&=ma)\\24&=(4+6+2)a\\a&=2.0\text{ m }{{\text{s}}^{{-2}}}\end{aligned}

If we consider blocks B and C as a combined 8 kg mass, then \begin{aligned}({{F}_{{net}}}&=ma)\\{{F}_{{AB}}}&=(6+2)(2.0)\\&=16\text{ N}\end{aligned}

If we consider block A by itself as a 4 kg mass, then \displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\24-{{F}_{{BA}}}&=(4)(2.0)\\{{F}_{{BA}}}&=16\text{ N}\end{aligned}

b)

Considering block C by itself as a 2 kg mass, \displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\{{F}_{{BC}}}&=(2)(2.0)\\&=4.0\text{ N}\end{aligned}

Alternatively, we can consider blocks A and B as a combined 10 kg mass. In which case \displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\24-{{F}_{{CB}}}&=(4+6)(2)\\{{F}_{{CB}}}&=4.0\text{ N}\end{aligned}

For the sake of completeness, the full picture is presented below. You can check that the acceleration always comes up to be 2.0 m s-2, no matter which mass you choose to apply your N2L on, as long as you correctly gather the forces that make up Fnet.

Video Example

Man in Elevator

Video Examples with Forces in 2D

Block on Slope

Pendulum in Accelerating Train

Video Examples with Multiple Bodies

Horizontal Stack

Atwood’s Machine

Vertical Stack