Lesson 3 of 5Dynamics

Unit 1 · Lesson 3 of 5

Connected Objects and Frames of Reference

Solving for the acceleration and tension when objects are joined by strings and pulleys, and understanding why a scale in an elevator or a passenger in a braking bus seems to feel forces that are not really there.

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Key things to remember

1Treat objects moving together as a system: $a = \frac{F _{n e t, e x t er na l}}{m _{t o t a l}};$ tension is internal and cancels.
2Find tension from one object's free-body diagram, for example $F_{T} = m_{2} (g - a)$ for a hanging mass accelerating down.
3Atwood machine: $a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}$ .
4An inertial frame is not accelerating; Newton's laws hold there using only real forces.
5In an accelerating elevator the scale reads $F_{N} = m (g + a)$ with up as positive; in free fall it reads zero.

Elevators, tow trucks and dumbwaiters all involve several objects that move together because a cable or a hitch links them. You could analyze each object alone, but the objects are not independent: a rope pulls one forward and the other back. This topic gives you two complementary tools, the system approach and separate free-body diagrams, and then asks a deeper question: does the answer depend on who is watching?

Connected objects: system and individual diagrams

Fig 1The same tension pulls the cart toward the pulley and pulls the hanging mass up. Each object gets its own free-body diagram.

Assume an ideal string (no mass, does not stretch) and an ideal pulley (no mass, no friction). Then the tension is the same all along the string, and both objects have the same size of acceleration. Treat everything that moves together as one system: the tension forces are internal, pulling one part forward and another back by equal amounts, so they cancel. Only external forces appear in $F_{n e t} = m_{t o t a l} a$ . Once you have $a$ , go back to the free-body diagram of one object to find the tension.

Cart on frictionless table: a = \frac{m _{2} g}{m _{1} + m _{2}} Atwood machine: a = \frac{( m _{2} - m _{1} ) g}{m _{1} + m _{2}}

m2 is the hanging mass (or the heavier mass in an Atwood machine). The denominator is always the total mass being accelerated.

Tension is not the weight of the hanging mass

If the hanging mass accelerates downward, the string must pull up on it with less than its weight, or it would not accelerate down:

F_{T} = m_{2} (g - a)

. Only when the system is at rest or moving at constant velocity does the tension equal

m_{2} g

Choose a positive direction that follows the motion around the pulley, for example toward the pulley on the table and downward for the hanging mass.
Find the net external force on the whole system in that direction: the driving weight minus any friction.
Divide by the total mass to get the acceleration.
Isolate one object, draw its free-body diagram, and apply $F_{n e t} = ma$ to find the tension.
Check with the other object's diagram. Both must give the same tension.

Inertial and non-inertial frames of reference

AInertial frame

At rest or moving at constant velocity
Newton's laws work using only real forces
Every force has an agent: Earth, a rope, a surface
Example: the ground, for everyday problems

BNon-inertial frame

Accelerating: speeding up, slowing down or turning
Objects seem to accelerate with no real force causing it
Observers invent apparent (fictitious) forces to explain this
Example: a braking bus, an accelerating elevator, a turning car

Fig 2The scale reads the normal force, not the weight. They differ whenever the elevator accelerates.

Elevator motion	Acceleration	Scale reading	Feels
At rest or constant velocity	Zero	$F_{N} = m g$	Normal
Speeding up going up, or slowing down going down	Upward	$F_{N} = m (g + a)$	Heavier
Speeding up going down, or slowing down going up	Downward	$F_{N} = m (g - a)$	Lighter
Cable cut: free fall	$9.81 m/s^{2}$ downward	$F_{N} = 0$	Weightless

Apparent forces in accelerating vehicles

When a bus brakes, you lurch forward. From the road (an inertial frame) nothing pushed you: you kept moving while the bus slowed. From inside the bus it feels like a forward force. A pendulum hanging from the ceiling of a car that speeds up swings back and settles at an angle where

tan θ = a / g

, which lets you measure the car's acceleration. A crate on a flatbed truck stays put only while static friction can supply

ma

; if the truck accelerates harder than

μ_{s} g

, the crate slides backward relative to the truck.

Pick one frame and stay in it

In an inertial frame, draw only real forces. Never add a forward 'braking force' on the passenger or an outward force on a turning car. Almost every problem in this course is solved in the ground frame, where Newton's laws need no correction.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

2.0 kg

cart on a frictionless table is tied by a string over a pulley to a

0.50 kg

hanging mass. Find the acceleration and the tension.

Example 2

An Atwood machine has a

3.0 kg

mass and a

5.0 kg

mass on either side of an ideal pulley. Find the acceleration and the tension.

Example 3

4.0 kg

block on a table is tied over a pulley to a

3.0 kg

hanging mass. The coefficient of kinetic friction between block and table is

0.25

. Find the acceleration and the tension.

Example 4

60 kg

student stands on a bathroom scale in an elevator. What does the scale read when the elevator accelerates upward at

1.5 m/s^{2}

, and when it accelerates downward at

1.5 m/s^{2}

Where marks get lost

Setting the tension equal to the hanging mass's weight. That is only true when nothing accelerates.
Dividing the net force by just one mass instead of the total mass of everything that accelerates together.
Including tension in the system's net force. Inside a system the tension forces cancel in pairs.
Thinking an elevator moving downward must make the scale read less. The reading depends on the direction of the acceleration, not the velocity.
Adding an apparent force, such as a forward push on a braking passenger, to a free-body diagram drawn in the ground frame.

Flashcards 8

Ideal string and pulley

System approach

Atwood machine acceleration

Inertial frame of reference

+ 4 more

Practice

An Atwood machine holds

2.0 kg

and

2.5 kg

masses over an ideal pulley. What is the acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

For the same Atwood machine (

2.0 kg

and

2.5 kg

a = 1.09 m/s^{2}

), what is the tension in the string, in newtons? Give three significant figures.

1.5 kg

cart on a frictionless track is pulled by a string over a pulley attached to a

0.25 kg

hanging mass. What is the acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

55 kg

person stands on a scale in an elevator that accelerates downward at

2.0 m/s^{2}

. What does the scale read, in newtons? Give three significant figures.

A pendulum hanging from the roof of an accelerating car settles at 8.0° from the vertical. What is the car's acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

30 N

horizontal push is applied to a

4.0 kg

block, which pushes a

2.0 kg

block in front of it across a frictionless floor. What force does the

4.0 kg

block exert on the

2.0 kg

block?

An elevator's cable snaps and it falls freely. What does a bathroom scale inside it read?

Which of these is an inertial frame of reference, to a good approximation?

What name is given to a force, like the forward 'push' felt in a braking bus, that has no physical agent and appears only in an accelerating frame?

A crate rests on the flat bed of a truck. The coefficient of static friction between them is

0.35

. What is the largest acceleration the truck can have without the crate sliding, in

m/s^{2}

? Give three significant figures.

m/s²

Order these elevator situations for a person on a scale, from the highest scale reading to the lowest.

1Falling freely after the cable breaks
2Moving down and speeding up at 2 m/s²
3Moving up at a constant 5 m/s
4Moving up and speeding up at 3 m/s²
5Moving down and slowing down at 1 m/s²

Match each situation to the force that the free-body diagram needs.

Cart pulled across a table by a hanging mass

Crate staying put on an accelerating truck

Person on a scale in an elevator

Passenger lurching forward in a braking bus (ground frame)

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