01 · Stronger drive
For the same car and trailer, driving force increases to 2400 N. Find a and Q.
Hint
Use the combined equation, then the trailer.
Worked solution
a = (2400 − 240)/1200 = 1.8 m/s². Q = 300 × 1.8 + 60 = 600 N: tension.
Understand · explore · practise
Tow-bars, tension and thrust during braking
Use signed coupling forces to distinguish a tow-bar in tension from compression, solve driving and braking models and compare rods with strings.
Before you startConnected particles, Newton’s second law and signed acceleration.
01 / A rigid rod can pull or push
Choose a signed coupling force before deciding its type.
A light rigid tow-bar keeps a fixed separation and can transmit tension or thrust.
A positive force toward the car on the trailer is a pull: the bar is in tension. A negative result means the force points away from the car: the bar is in compression and exerts a thrust. Unlike a string, a rigid rod can push. The light model neglects the bar’s mass.
Driving, coasting and brakingExplore
A 900 kg car tows a 300 kg trailer to the right with a light rigid horizontal tow-bar. Resistances are 180 N on the car and 60 N on the trailer, both leftward while moving right. Any brake force acts on the car. Q is the signed force of the bar on the trailer, positive rightward.
02 / Combine the car and trailer to find acceleration
The coupling forces are internal.
A 900 kg car and 300 kg trailer move right. Driving force is 1200 N; resistances are 180 N and 60 N.Worked example
1200 − 180 − 60 = 1200a
Take right positive.
a = 0.8 m/s²
Both bodies have the same acceleration.
Trailer: Q − 60 = 300 × 0.8
The engine force does not act directly on the trailer.
Q = 300 N
Positive Q pulls the trailer toward the car: tension.
Watch: a signed coupling force changes from tension to thrust
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Car check
Check the original 1200 N drive using the car alone.
Hint
The bar pulls backward on the car with magnitude 300 N.
Worked solution
1200 − 180 − 300 = 720 = 900 × 0.8 N.
03 / Constant speed can still require coupling tension
Zero resultant does not mean every force is zero.
03 · Steady speed
Find the driving force and bar tension when the same system moves right at constant speed.
Hint
Set a = 0 for the whole system and trailer.
Worked solution
Driving force must be 180 + 60 = 240 N. Trailer: Q − 60 = 0, so Q = 60 N: tension.
04 · Coasting
Switch off the drive without applying the brakes. Find a and Q while both still move right.
Hint
The only external horizontal forces are the two resistances.
Worked solution
a = −240/1200 = −0.2 m/s². Q = 300(−0.2) + 60 = 0 N. Here each body’s resistance per kilogram happens to be equal, so no coupling force is required.
05 · General coasting claim
Does switching off the engine always make the tow-bar force zero?
Hint
Compare resistance divided by mass for the two bodies.
Worked solution
No. Q = 0 in question 4 because 180/900 = 60/300. Different free accelerations require a coupling force to preserve the rigid separation.
04 / Braking the car may compress the tow-bar
Resistance still opposes velocity while acceleration points backward.
No drive; an extra 960 N braking force acts on the car while the system moves right.Worked example
−960 − 180 − 60 = 1200a
Keep the same right-positive axis.
a = −1 m/s²
This is deceleration while velocity remains rightward.
Q − 60 = 300(−1)
Use the trailer to find the coupling.
Q = −240 N
The bar pushes the trailer left with thrust 240 N and pushes the car right with 240 N.
06 · Gentler braking
Replace 960 N by 480 N, with no drive. Find a and the bar force.
Hint
First sum the external forces.
Worked solution
a = −720/1200 = −0.6 m/s². Q = 300(−0.6) + 60 = −120 N: thrust 120 N.
07 · Car under braking
Check the car equation in the 960 N braking example.
Hint
The compressed bar pushes the car forward.
Worked solution
−960 − 180 + 240 = −900 = 900(−1) N.
08 · Direction audit
While braking to a stop from rightward motion, should the resistances reverse to the right?
Hint
Resistance opposes velocity, not acceleration.
Worked solution
No. Before stopping the velocity is still rightward, so both stated resistances remain leftward. The stage ends at rest; any later motion needs a fresh model.
05 / Use a known thrust to infer the brake force
State whether the given coupling force is a pull or a push.
09 · Unknown brake
In the same model, the bar has thrust 180 N, with no drive. Find acceleration and brake magnitude B on the car.
Hint
Thrust means Q = −180 N.
Worked solution
Trailer: −180 − 60 = 300a gives a = −0.8 m/s². Whole: −B − 240 = 1200(−0.8), so B = 720 N.
10 · Unknown drive
The bar has tension 450 N with no brakes. Find the driving force D.
Hint
Tension means Q = +450 N.
Worked solution
450 − 60 = 300a gives a = 1.3 m/s². D − 240 = 1200 × 1.3 gives D = 1800 N.
11 · Different trailer resistance
Keep masses and car resistance, but set trailer resistance to 120 N. With no drive or braking, find Q.
Hint
Use total resistance 300 N.
Worked solution
a = −300/1200 = −0.25 m/s². Q = 300(−0.25) + 120 = 45 N: tension, even while the system slows.
06 / Stop the braking equations when the speed reaches zero
Constant forces give a constant-acceleration stage only over its valid interval.
12 · Stopping
At the start of the 960 N braking stage the speed is 15 m/s rightward. Find stopping time and distance.
Hint
Use a = −1 m/s² until v = 0.
Worked solution
0 = 15 − t gives t = 15 s. Distance = (15 + 0) × 15/2 = 112.5 m.
13 · An earlier instant
Find speed and distance 6 s into the same braking stage.
Hint
Check 6 < 15 before using the stage.
Worked solution
v = 15 − 6 = 9 m/s. s = 15 × 6 − ½ × 6² = 72 m.
07 / Replacing the rod with a string changes the model
A string cannot supply the negative Q needed during these braking cases.
14 · Rope instead
Could a taut tow-rope provide the −240 N force on the trailer in the braking example?
Hint
A rope can pull the trailer toward the car but cannot push it away.
Worked solution
No. The calculated force requires compression. A rope would become slack as the more strongly decelerating car is approached by the trailer; solve separate motions once the taut condition fails, until another event changes the model.
Report “thrust 240 N”, not a negative magnitude. The sign belongs to the chosen directional force variable Q. Draw opposite coupling forces on the two bodies and check that they cancel only in the combined system.
08 / Use a sign to diagnose tension or thrust
The whole system gives acceleration; one body gives the coupling force.
Fix the axis, draw the external forces and find the shared acceleration. Then calculate the signed bar force. Interpret its direction geometrically, keep resistance opposite motion, and restrict braking calculations to the interval before stopping.
Section 1 of 8 · A rigid rod can pull or push