Coaching Centre
Constant net force and connected bodies
Motion under a constant net force described with words, graphs and vectors; Newton's second law for connected bodies (cart and hanging mass, Atwood machine, towing); tension in strings; and the F = ma practical with constant total mass.
A lift is a box on one end of a cable and a heavy counterweight on the other, running over a pulley at the top. Because the two sides nearly balance, the motor only has to supply the difference. The physics is the same as a toy you can build on a desk with two masses and a pulley.
Newton's second law, $\vec F_{net} = m\vec a$, is most useful when the net force is constant. Then the acceleration is constant, and everything from Module 1 applies: the equations of motion and straight-line velocity–time graphs. This page also shows how to handle several objects joined by strings.
What a constant net force does
- In words: a constant net force gives a constant acceleration in the direction of the force. If the force is along the motion, the object speeds up steadily; against the motion, it slows steadily; at an angle, the path curves.
- In graphs: $a$–$t$ is a horizontal line, $v$–$t$ is a straight line with gradient $a$, and $s$–$t$ is a parabola.
- In vectors: each second, the velocity vector changes by the same vector $\vec a\,\Delta t$, which points the same way as $\vec F_{net}$.
The second law also tells you how acceleration depends on each variable. With the mass fixed, $a \propto F_{net}$: double the force, double the acceleration. With the force fixed, $a \propto \dfrac{1}{m}$: double the mass, half the acceleration.
Connected bodies: the system method
When objects are joined by a light, inextensible string, they move together with the same size of acceleration. Two steps solve almost every problem:
- Whole system: treat everything as one object. Internal forces such as tension cancel. $a = \dfrac{\text{net external force}}{\text{total mass}}$.
- One object: isolate one body and apply $\Sigma F = ma$ to it to find the tension.
A 0.80 kg cart on a smooth bench is joined by a string over a light pulley to a 0.10 kg hanging mass. Find the acceleration and the tension.
Masses of 2.0 kg and 1.5 kg hang on either side of a light, frictionless pulley. Find the acceleration and tension, and the speed after the masses have moved 0.60 m from rest.
A 1500 kg car tows a 750 kg trailer and accelerates at 1.2 m s⁻². Rolling resistance is 300 N on the car and 150 N on the trailer. Find the driving force and the tension in the tow bar.
The F = ma practical
A student's graph of acceleration against hanger weight has a gradient of 1.12 kg⁻¹ and an $x$-intercept of 0.06 N. Interpret both.
- With a 0.80 kg cart and a 0.10 kg hanger, press Release. The acceleration is 1.09 m/s² and the tension is 0.871 N, less than the hanger's 0.98 N weight.
- Press Record run, then Move 20 g from cart to hanger and record again, a few times. The total stays at 0.90 kg while the acceleration rises in equal steps: 1.09, 1.31, 1.52 m/s².
- Add friction μk = 0.10 to the cart with the 0.80 kg + 0.10 kg setup: the acceleration drops to 0.22 m/s².
- Switch to the Atwood machine with 2.0 kg and 1.5 kg: $a$ = 1.40 m/s² and $T$ = 16.8 N. Make the masses equal and nothing accelerates.
Test yourself
A 3.0 kg block on a frictionless table is joined over a pulley to a 1.0 kg hanging mass. The acceleration of the system is
$a = \dfrac{1.0 \times 9.8}{4.0} = 2.45$ m s⁻².
In the question above, the tension in the string is
For the block: $T = 3.0 \times 2.45 = 7.35$ N.
An object has a constant net force acting on it in its direction of motion. Its velocity–time graph is
Constant force means constant acceleration, which is the constant gradient of a $v$–$t$ graph.
The net force on a trolley is doubled while its mass is halved. Its acceleration
$a = F/m$: $(2F)/(m/2) = 4F/m$.
Label the forces on the cart and the hanger.
Order the steps to find the tension between two connected bodies.
In a cart-and-hanger system, the tension in the string is than the hanger's weight while it accelerates. To test $a \propto F$ fairly, the total of the system must stay constant, so masses are moved from the to the hanger. A graph of acceleration against net force should be a straight line through the .
A student investigates $F = ma$ by adding 50 g masses one at a time to a hanger attached to a 0.50 kg cart, without moving any mass from the cart. The track is level and friction is not compensated. Evaluate this method, and explain the shape of the acceleration against hanger-weight graph the student would get.
The method is not a valid test of $a \propto F$, because two variables change at once.
Adding mass to the hanger increases the net force, but it also increases the total mass being accelerated. The acceleration is $a = \dfrac{mg}{M + m}$, so as $m$ grows the denominator grows too, and the graph of $a$ against $mg$ curves, with a gradient that gets smaller. A curve could wrongly suggest that acceleration is not proportional to force. Moving masses from the cart to the hanger keeps $M + m$ constant, so the graph should be a straight line with gradient $1/(M + m)$.
Because friction is not compensated, the cart will not move until the hanger's weight exceeds the friction, so the graph will not pass through the origin: it will cut the force axis at a positive value. All accelerations will also be lower than predicted, a systematic error. Tilting the track until the cart rolls at constant speed after a push removes this error.
With both changes the method would validly test the relationship. Repeating each trial three times would also improve reliability.