Review: forces and Newton's laws
Use this page to pull the whole unit together. Read the summary, then work through the questions without looking at the solutions. If one stumps you, go back to the lesson listed next to it.
Key ideas of the unit
| Idea | What to remember | Lesson |
|---|---|---|
| Force | A push or pull by one object on another; a vector measured in newtons, \(1\ \text{N} = 1\ \text{kg}\cdot\text{m/s}^2\). | 1 |
| Free-body diagram | One object, only the forces acting on it, drawn as arrows from its center. | 1 |
| First law | No net force means constant velocity (possibly zero). Mass measures inertia. | 2 |
| Equilibrium | \(\sum F_x = 0\) and \(\sum F_y = 0\). | 2 |
| Second law | \(\sum \vec F = m\vec a\); the acceleration points along the net force. | 3 |
| Third law | \(\vec F_{\text{A on B}} = -\vec F_{\text{B on A}}\); the pair acts on different objects. | 4 |
| Weight | \(F_g = mg\), with \(g = 9.81\ \text{m/s}^2\) at Earth's surface. | 5 |
| Normal force | Perpendicular to the surface; find it from \(\sum F_y = 0\). On a ramp with no other forces, \(F_n = mg\cos\theta\). | 5 |
| Friction | Static: up to \(\mu_s F_n\). Kinetic: \(\mu_k F_n\). | 5 |
| Air resistance | Grows with speed; terminal speed when it equals the weight. | 6 |
Concept questions
Question 1
A spacecraft far from any star or planet has its engines off. What does it do?
- It slows down and eventually stops.
- It speeds up.
- It keeps moving in a straight line at constant speed.
- It starts to move in a circle.
Show solution
With no net force, Newton's first law says its velocity stays constant. Engines are only needed to change its speed or direction.
Question 2
A cup sits at rest on a table. What can you conclude?
- No forces act on the cup.
- The forces on the cup add up to zero.
- Only gravity acts on the cup.
- The table pushes up harder than gravity pulls down.
Show solution
Being at rest means zero acceleration, so the net force is zero. Gravity and the normal force both act, and they balance.
Question 3
You jump from a small rowboat onto a dock, and the boat drifts away from the dock. Which statement explains this?
- The dock pushes the boat away.
- Your weight pushes the boat backward.
- The water pulls the boat away once you leave it.
- You push the boat backward, and the boat pushes you forward with an equal force.
Show solution
Your legs push backward on the boat; by the third law the boat pushes you forward. The boat's push on you launches you to the dock, and your push on the boat sends it the other way.
Question 4
A crate slides across a floor and slows down. Which force is responsible, and what kind is it?
- Kinetic friction, a contact force
- Static friction, a contact force
- Gravity, a field force
- The normal force, a field force
Show solution
The crate is sliding, so the friction on it is kinetic. Friction needs contact between the surfaces.
Question 5
Two balls of the same size, one steel and one rubber, are dropped from a high tower. Air resistance is not negligible. Which reaches the higher terminal speed?
- The rubber ball, because it is lighter
- They reach the same terminal speed
- The steel ball, because air resistance must grow larger to balance its weight
- Neither, because objects in air never reach terminal speed
Show solution
Terminal speed is reached when air resistance equals the weight. Same size means the same air resistance at the same speed, so the heavier steel ball has to fall faster before air resistance balances it.
Problems
Problem 1 · Lesson 3
A 2.0 kg book is pushed across a table with a horizontal force of 8.0 N. Friction on the book is 3.0 N. Find its acceleration.
Show solution
\(\sum F = 8.0 - 3.0 = 5.0\ \text{N}\), so \(a = 5.0 / 2.0 = 2.5\ \text{m/s}^2\) in the direction of the push.
Problem 2 · Lesson 2
A hot-air balloon is pushed by the wind with 90 N toward the east, while its burner gives a net upward force of 120 N. What is the size of the net force on the balloon?
Show solution
\(F_{\text{net}} = \sqrt{90^2 + 120^2} = 150\ \text{N}\), at \(\tan^{-1}(90/120) = 36.9^\circ\) east of straight up.
Problem 3 · Lesson 2
A 200 N traffic light hangs at rest from two cables. Each cable makes an angle of 30.0° below the horizontal, one on each side. Find the tension in each cable.
Show solution
By symmetry the tensions are equal. The horizontal parts cancel. Vertically, \(2T\sin 30.0^\circ = 200\ \text{N}\), so \(T = \dfrac{200}{2(0.500)} = 200\ \text{N}\).
Each cable carries as much as the whole weight. The flatter the cables, the larger the tension.
Problem 4 · Lesson 5
What does a 75.0 kg astronaut weigh on the Moon, where \(g = 1.62\ \text{m/s}^2\)?
Show solution
\(F_g = (75.0)(1.62) = 122\ \text{N}\)
Problem 5 · Lesson 5
A 10.0 kg box sits on the floor. You push straight down on its top with 30.0 N. What is the normal force from the floor?
Show solution
\(\sum F_y = 0\): \(F_n - mg - 30.0 = 0\), so \(F_n = 98.1 + 30.0 = 128\ \text{N}\).
Problem 6 · Lesson 5
A book on a tilting board starts to slide when the board reaches 22.0°. What is the coefficient of static friction?
Show solution
\(\mu_s = \tan 22.0^\circ = 0.40\)
Problem 7 · Lesson 5
A 50.0 kg crate is pushed across a level floor with a horizontal force of 200 N. The coefficient of kinetic friction is 0.30. Find its acceleration.
Show solution
\(F_k = 0.30 \times (50.0)(9.81) = 147\ \text{N}\). \(a = \dfrac{200 - 147}{50.0} = 1.06\ \text{m/s}^2\).
Problem 8 · Lessons 3 and 5
A 60.0 kg student stands on a bathroom scale in an elevator that is speeding up on its way up with an acceleration of 1.50 m/s². What does the scale read?
Show solution
The scale reads the normal force. Take up as positive: \(F_n - mg = ma\), so \(F_n = m(g + a) = 60.0(9.81 + 1.50) = 679\ \text{N}\).
That is more than the student's weight of 589 N, which is why you feel heavier when an elevator starts upward.
Multi-step problems
Problem 9 · Lessons 3, 5 and motion
A 1500 kg car traveling at 25.0 m/s locks its brakes on a dry road and skids to a stop. The coefficient of kinetic friction between the tires and the road is 0.70. How far does it skid? Try it before you open the steps.
- Friction is the only horizontal force. \(F_n = mg\), so \(F_k = \mu_k mg\).
- Acceleration. \(a = \dfrac{F_k}{m} = \mu_k g = 0.70 \times 9.81 = 6.87\ \text{m/s}^2\), opposite the motion. The mass cancels.
- Distance. With \(v_f = 0\): \(v_f^2 = v_i^2 - 2a\Delta x\), so $$\Delta x = \frac{v_i^2}{2a} = \frac{(25.0)^2}{2(6.87)} = 45.5\ \text{m}$$
The car skids about 45.5 m, whatever its mass.
Problem 10 · Lesson 6
A child pulls a 20.0 kg sled across level snow at constant velocity with a rope at 35.0° above the horizontal. The coefficient of kinetic friction is 0.15. What is the tension in the rope?
- Constant velocity means equilibrium in both directions.
- Vertical. \(F_n = mg - F\sin 35.0^\circ\). The rope lifts part of the weight.
- Horizontal. \(F\cos 35.0^\circ = \mu_k F_n = \mu_k(mg - F\sin 35.0^\circ)\).
- Solve for \(F\). $$F = \frac{\mu_k mg}{\cos 35.0^\circ + \mu_k\sin 35.0^\circ} = \frac{0.15(196)}{0.819 + 0.086} = 32.5\ \text{N}$$
\(F = 32.5\ \text{N}\)
Ready for the next unit?
- If you could do most of these without the solutions, move on to Work.
- If friction problems were hard, try the forces and friction lab and redo Problems 7 and 10.
Finished?
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