Tow-bars, systems & stage changes | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Use the whole system for acceleration, then one body for tension

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Whole system for acceleration; one body for the internal force.

Internal tension appears with opposite directions on the two bodies, so it cancels when the equations are added. This is the equal and opposite interaction pair from Newton's third law. External forces determine the acceleration of the whole system.

Worked example

A car tows a trailer

A 900 kg car tows a 300 kg trailer. The driving force is 1600 N. Resistances on the car and trailer are 200 N and 100 N.

1600-200-100=1300

1300=1200a

a=\frac{13}{12}\text{ m s}^{-2}

Now isolate the trailer:

T-100=300\left(\frac{13}{12}\right)

T=425\text{ N}

Common mistake

When using the whole system, include every external resistance but do not include the internal tension or thrust.

A rod may be in tension or in thrust

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A rod can pull or push. A changed contact or tautness condition starts a new force model.

A light rope can pull, so its internal force is tension.
A rigid tow-bar keeps the distance between the bodies fixed. It can pull in tension or push in thrust.
If you assume tension and obtain a negative value, the bar is actually in thrust with that positive magnitude.
State the force type and magnitude in the final answer.

Worked example

Interpret a negative internal force

Suppose the one-body equation for a trailer is $T-500=300(1)$ , giving $T=800\text{ N}$ . The bar is in tension. If instead the equation produced $T=-200\text{ N}$ , the correct statement would be “thrust of 200 N”.

When contact or a string condition changes, start a new model

If a hanging particle reaches the ground, the string may become slack. A slack string cannot pull, so its tension is zero. The other particle keeps its current velocity at that instant because the model contains no instantaneous impulse on it, but its acceleration changes with the new force diagram.

Solve the first stage with the connected-particle model.
Use kinematics to find the velocity at the event.
State what changes physically at the event.
Redraw forces and find the new acceleration.
Use the event velocity as the next stage's initial velocity.

Examiner note

In a current paper, once the hanging particle reached the ground the mark scheme required a new Newton equation for the particle on the plane and then a new constant-acceleration calculation. Carrying the old tension or acceleration into the second stage loses the model. (9709/42/O/N/24 Q7(b))

Your turnIndependent check [7 marks]

A 2 kg particle A lies on a rough horizontal table with coefficient of friction 0.2. It is connected over a smooth pulley to a hanging 1 kg particle B, initially 0.8 m above the floor. The system starts from rest. Find A's total distance travelled before it first comes to rest.

Show worked answer

Stage 1: the friction on A is $0.2(20)=4\text{ N}$ .

10-4=(2+1)a

a=2\text{ m s}^{-2}

v^2=0+2(2)(0.8)=3.2

Stage 2: B is on the floor, the string is slack and $T=0$ . Friction decelerates A:

-4=2a

a=-2\text{ m s}^{-2}

0=3.2+2(-2)s

s=0.8\text{ m}

s_{\mathrm{total}}=0.8+0.8

s_{\mathrm{total}}=1.6\text{ m}

Answer

a_1=2\text{ m s}^{-2}

v^2=3.2

a_2=-2\text{ m s}^{-2}

s_{\mathrm{total}}=1.6\text{ m}

Use the whole system for acceleration, then one body for tension

Swipe left or right to see the whole diagram →

Whole system for acceleration; one body for the internal force.

Worked example

A car tows a trailer

A 900 kg car tows a 300 kg trailer. The driving force is 1600 N. Resistances on the car and trailer are 200 N and 100 N.

1600-200-100=1300

1300=1200a

a=\frac{13}{12}\text{ m s}^{-2}

Now isolate the trailer:

T-100=300\left(\frac{13}{12}\right)

T=425\text{ N}

Common mistake

When using the whole system, include every external resistance but do not include the internal tension or thrust.

A rod may be in tension or in thrust

Swipe left or right to see the whole diagram →

A rod can pull or push. A changed contact or tautness condition starts a new force model.

A light rope can pull, so its internal force is tension.
A rigid tow-bar keeps the distance between the bodies fixed. It can pull in tension or push in thrust.
If you assume tension and obtain a negative value, the bar is actually in thrust with that positive magnitude.
State the force type and magnitude in the final answer.

Worked example

Interpret a negative internal force

When contact or a string condition changes, start a new model

Solve the first stage with the connected-particle model.
Use kinematics to find the velocity at the event.
State what changes physically at the event.
Redraw forces and find the new acceleration.
Use the event velocity as the next stage's initial velocity.

Examiner note

Your turnIndependent check [7 marks]

Show worked answer

Stage 1: the friction on A is $0.2(20)=4\text{ N}$ .

10-4=(2+1)a

a=2\text{ m s}^{-2}

v^2=0+2(2)(0.8)=3.2

Stage 2: B is on the floor, the string is slack and $T=0$ . Friction decelerates A:

-4=2a

a=-2\text{ m s}^{-2}

0=3.2+2(-2)s

s=0.8\text{ m}

s_{\mathrm{total}}=0.8+0.8

s_{\mathrm{total}}=1.6\text{ m}

Answer

a_1=2\text{ m s}^{-2}

v^2=3.2

a_2=-2\text{ m s}^{-2}

s_{\mathrm{total}}=1.6\text{ m}