Free body diagrams are a technique that helps visualize the force
vectors used to solve Newton's second law.
A free body diagram shows all the force vectors on an object.
Mass is typically written inside a box, with force vectors pointed away from the box.
Example: Draw a free body diagram for a 83 kg person holding onto a rope in a game of "Tug of
war". The rope is pulling them with a force of 520 N left. The person's feet are countering that with 540 N to the
right.
solution
Example: Use the free body diagram to calculate the mass's acceleration.
solution
$$\sum F=ma$$
$$-200\,\mathrm{N}+300\,\mathrm{N}+500\,\mathrm{N}=ma$$
$$600\,\mathrm{N}=(2\,\mathrm{kg})a$$
$$300 \mathrm{\tfrac{m}{s^2}}=a$$
Example: The book The Martian by Andy Weir (mass =
0.09 kg) is resting on a table (acceleration = zero). The force of gravity produces a downwards force of 0.34 N.
What force is required to keep the book at rest on the table?
solution
$$\sum F=ma$$ $$F_{N} + F_{g} = ma$$
$$F_{N} - 0.34\, \mathrm{N} = (0.09\, \mathrm{kg})(0)$$
$$F_{N} - 0.34\,\mathrm{N} = 0$$
$$F_{N} = 0.34 \, \mathrm{N}$$
Example: Use the free body diagram to calculate the acceleration for both the horizontal and the
vertical.
solution
$$ \text{vertical}$$
$$\sum F=ma$$
$$22 \, \mathrm{N} - 85 \, \mathrm{N} = (16 \, \mathrm{kg})a$$
$$-63 \, \mathrm{N} = (16 \, \mathrm{kg})a$$
$$-3.9 \, \mathrm{\tfrac{m}{s^2}} = a$$
$$ \text{horizontal}$$
$$\sum F=ma$$
$$78 \, \mathrm{N}+55 \, \mathrm{N}-140 \, \mathrm{N} = (16 \, \mathrm{kg})a$$
$$-7 \, \mathrm{N} = (16 \, \mathrm{kg})a$$
$$-0.43 \, \mathrm{\tfrac{m}{s^2}} = a$$
Example: A red crate is falling, but a parachute is slowing its acceleration to only 2 m/s² down.
The force of gravity is 833 N down. The parachute provides 663 N up. Draw a free body diagram and use it to find
the mass of the crate.
solution
$$\sum F=ma$$ $$F_1+F_2=ma$$ $$-833+663 = m(2)$$
$$\frac{-170}{2} = \frac{m(2)}{2}$$
$$-85 \, \mathrm{kg} = m$$
A negative mass doesn't make sense. Where did we mess up?
Oh, the acceleration is negative, because it points down.
Mass is a measure of an object's inertia. Mass also determines the strength of
gravity. Because of gravity all objects are attracted to each other, but we mostly notice the attraction towards
the Earth because it is so large and so close.
There is a special word to describe the direction of gravity, down.
$$F_g=mg $$
\(F_g\) = the force of gravity, weight [N, newtons, kg m/s²]
vector \(m\) = mass [kg]
\(g\) = acceleration of gravity on Earth = 9.8 [m/s²]
vector
The force of gravity depends only on the mass of the object because on the surface of the Earth
acceleration from gravity is the same for all objects. If you aren't on the surface of the Earth,
there is a different way to calculate gravity.
Question: Why do feathers fall slower than bricks?
answer
Air friction produces a force that opposes motion.
Feathers have a large surface area compared to their small mass so they have more air friction.
The acceleration of 9.8 m/s² on the surface of the Earth is just an approximation. The
gravity of Earth changes a bit depending on where
you are.
Table: Comparative gravities in various cities around the world
Question: What factors might explain why the measured acceleration of gravity changes in
different locations on the surface of Earth?
answer
Gravity is slightly weaker at higher elevation because you are farther from the center of the Earth.
Near the equator gravity feels weaker because the Earth's rotation adds a centrifugal force.
Example: The Three-Body Problem by Cixin Liu has a mass of 0.44 kg. What force of gravity does
the book have?
solution
$$F_g=mg$$ $$F_g=(0.44 \, \mathrm{kg} )(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g=4.3 \, \mathrm{N}$$
Example: How much mass does the book Cat's Cradle by Kurt Vonnegut have if it feels a force of
gravity of 1.8 N?
solution
$$F_g=mg$$ $$\frac{F_g}{g}=m$$
$$\frac{1.8 \,\mathrm{N}}{9.8\, \mathrm{\tfrac{m}{s^2}}}=m$$
$$0.18 \, \mathrm{kg}=m$$
Weight and Mass
Another word for the force of gravity is weight. An object on the Moon would weigh less than it does on Earth because of
the lower gravity, but it would still have the same mass.
$$F_g = \mathrm{weight}$$
Earth's gravity does extend into space, but it decreases with distance. It is about ~90% for astronauts in orbit
around the Earth, but they don't notice any gravity because they are in a freefall.
Freefall means that you are just letting gravity accelerate you without any opposing forces. To
keep from falling we are careful to always counter the force of gravity. This can be done with a parachute, or a
jet pack, or just the ground.
Question: Ducks: Two Years in the Oil Sands by Kate Beaton has a mass of 1.14 kg.
How does its mass and weight change in a freefall on Earth?
answer
An object's mass doesn't change when it is falling.
Weight just means the force of gravity, which also doesn't change in a short freefall.
Example: The hardcover version of Seveneves by Neal Stephenson has a mass of 0.95 kg.
What is the force of gravity felt by the book on Earth? What about on the Moon?
Local Massive Objects Surface Gravity
name
g (m/s²)
Sun
275
Mercury
3.7
Venus
8.9
Earth
9.8
Moon
1.6
Mars
3.7
Jupiter
25.8
Saturn
10.4
Uranus
8.7
Neptune
11.2
solution
Weight and force of gravity mean the same thing. A planet's gravity field determines your weight, but not
your mass.
Pounds are a unit of force and kilograms are a unit of mass.
You can't convert directly between them because they are different concepts, but you can use the force of
gravity equation to find a conversion that works for only Earth's surface.
Example: Uprooted by Naomi Novik is resting on a table. The shipping weight is 1.2 pounds.
What is the book's mass in kilograms on Earth? On Mars?
solution
Mass doesn't depend on gravity, so it's the same everywhere.
$$ 1.2\,\mathrm{lbs} \left( \frac{1\,\mathrm{kg}}{2.2\,\mathrm{lbs}} \right)= 0.\overline{54}\,\mathrm{kg}$$
$$ m = 0. \overline{54} \, \mathrm{kg} $$
What is the weight of the book in Newtons on Earth? On Mars?
solution
$$\text{weight on Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(9.8)$$ $$F_{g}=5.35\, \mathrm{N}$$
$$\text{weight on Mars}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(3.711)$$ $$F_{g}=2.02\, \mathrm{N}$$
The Normal Force
Typically a normal force will balance the force of gravity to keep an object from accelerating up or down.
A normal force occurs when two objects are in contact. It is perpendicular to the point of
contact. A normal force prevents objects from passing through each other.
A normal force will scale to a value that will keep the net force and acceleration zero.
Normal forces come from the combined effect of electromagnetic forces and the Pauli exclusion principle.
The electromagnetic force allows chemical bonds to
form. These bonds give solid matter its rigid structure, which is required for normal forces.
At the atomic scale, particles can't pass through each other primarily because of a quantum mechanical effect
called the Pauli exclusion principle . The
Pauli exclusion principle is mostly responsible for keeping particles, like electrons, separate.
Press E to activate the mass. Then press WASD to apply forces to the mass.
Imagine that gravity is pointed towards the bottom of the page.
Question: What force keeps the mass from exiting the screen?
answer
The normal force.
Question: Why does pressing A and D at the same time do nothing?
answer
The left and right force cancel each other out.
Example: You are accelerating up in an elevator at 2 m/s². If your mass is 100 kg, what
is the normal force you feel from the elevator?
solution
$$F_{g}=mg$$ $$F_{g}=(100)(9.8)$$ $$F_{g}=980\, \mathrm{N}$$
$$\sum F=ma$$ $$F_{N}-F_{g}=ma$$ $$F_{N} - 980=(100)(2)$$ $$F_{N}=1180\, \mathrm{N}$$
Example: A 20 kg box is at rest on a horizontal sidewalk. Find the force of gravity and the
normal force on the box.
solution
In the simple case of a flat horizontal surface with no vertical acceleration the force of gravity will always
be equal and opposite to the normal force.
Example: A 20 kg box is at rest on a steep sidewalk. The sidewalk is at an angle 20 degrees from
horizontal. Find the force of gravity and the normal force on the box. What is the acceleration of the box?
(assume no friction)
solution
When a person is walking a dog they are able to apply a force on the dog with a leash. They use the tension on the
leash to transfer that force from their hand to the dog.
Tension is the pulling force from a chain, string, or rope. Tension is useful for transferring a force over a
distance. In most situations, the tension is the same for both ends.
Example: A helium balloon is attached to a 0.5 g paper clip. If the balloon and paper clip are
accelerating up at 0.023 m/s², what is the tension on the paper clip from the balloon in Newtons?
solution
$$\sum F=ma$$ $$-F_{g} + T = ma$$ $$T = ma + F_{g}$$
$$T = ma + mg$$
$$T = (0.0005\, \mathrm{kg})(0.023\, \mathrm{\tfrac{m}{s^2}})+ (0.0005\, \mathrm{kg}) (9.8\,
\mathrm{\tfrac{m}{s^2}})$$
$$T = 0.0049115\, \mathrm{N}$$
Example: A 100 kg person is pulling a 10 kg crate with a rope (ignore the mass of the rope).
Both the person and the crate are accelerating to the left at 0.1 m/s². What force is the person producing in
order to accelerate to the left?
solution
The tension force is equal and opposite for the person and crate.
Example: You are walking your dog with the leash at a 45 degree angle down towards the dog.
Neither you nor the dog are accelerating. The dog is pulling on the leash forward with a force of 100 N.
Calculate the x part of the tension force on the leash. Then calculate the total tension force.
solution
$$\text{dog: horizontal}$$
$$\sum F=ma$$
$$-T_x + F_{\mathrm{dog}} = 0$$
$$-T_x + 100 \, \mathrm{N} = 0$$
$$T_x= 100 \,\mathrm{N}$$
The 100 N is only the x-part of the tension force vector. We can find the total tension force with the
Pythagorean theorem. At 45° the x and y parts of the force are the same.
Example: A 0.4 kg squirrel is pulling a 10 kg box with a string. The squirrel and box are
accelerating to the left at 0.5 m/s². What is the force of tension on the string? What force is the squirrel
producing in order to accelerate to the left?
solution
$$ \text{box}$$ $$\sum F=ma$$ $$F_{\mathrm{tension}}=(10)(-0.5)$$ $$F_{\mathrm{tension}}=-5\, \mathrm{N}$$
The tension force on the string is equal but opposite for the squirrel and box.
Example: Use the diagram to predict the acceleration of the masses.
Assume no friction, no air resistance, Earth gravity, and a massless rope.
hint
Both free body diagrams share the same tension and acceleration. Build two equations with Newton's second law
and solve a system of equations for T and a. While T and a are the same magnitude they have different
directions, so watch the sign of acceleration in particular.
This Khan academy video
covers this problem in more depth.
Example: Two blocks are attached to each other with a rope hanging over a wheel. Find the
acceleration of each body. Ignore friction, and assume the wheel and rope have a negligible mass.
solution
$$\text{10 kg block}$$
$$\sum F = ma$$
$$T-mg = ma$$
$$T-(10)(9.8) = ma$$
$$T-98 = 10a$$
$$T = 10a+98$$
$$\text{25 kg block}$$
$$\sum F = ma$$
$$T-mg = ma$$
$$T-(25)(9.8) = 25a$$
$$T-245 = 25a$$
$$T = 25a+245$$
We have two variables and two equations, this means we can plug one equation into the other. The Tension is
the same for each body. The accelerations are the same, but in opposite directions, so we need to make one
acceleration negative.
In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!
Unless a problem says otherwise, assume Earth gravity, no air resistance, no friction, and massless ropes.
Example: A 6 kg box has three horizontal forces on it: 18 N right, 10 N left, and 4 N right. What is the box's horizontal acceleration?
solution
Let right be positive.
$$\sum F_x = 18\,\mathrm{N} - 10\,\mathrm{N} + 4\,\mathrm{N}$$
$$\sum F_x = 12\,\mathrm{N}$$
$$\sum F=ma$$
$$a_x=\frac{\sum F_x}{m}$$
$$a_x=\frac{12\,\mathrm{N}}{6\,\mathrm{kg}}$$
$$a_x=2\,\mathrm{m/s^2}$$
The box accelerates to the right.
Example: A 2500 g textbook rests on a flat table next to a pencil and notebook. What are the force of gravity and the normal force on the book?
solution
$$2500\,\mathrm{g}=2.5\,\mathrm{kg}$$
$$F_g=mg$$
$$F_g=(2.5\,\mathrm{kg})(9.8\,\mathrm{m/s^2})$$
$$F_g=24.5\,\mathrm{N}$$
The book is not accelerating up or down, so the vertical forces cancel.
$$F_N-F_g=0$$
$$F_N=24.5\,\mathrm{N}$$
Example: An 80 kg person stands on a scale in an elevator accelerating upward at 150 cm/s². The elevator display says floor 12, but the scale reading comes from the normal force. What normal force does the floor apply to the person?
solution
Let up be positive.
$$150\,\mathrm{cm/s^2}=1.5\,\mathrm{m/s^2}$$
$$F_g=mg$$
$$F_g=(80)(9.8)$$
$$F_g=784\,\mathrm{N}$$
$$\sum F=ma$$
$$F_N-F_g=ma$$
$$F_N-784=(80)(1.5)$$
$$F_N=904\,\mathrm{N}$$
Example: A 60 kg person stands in an elevator accelerating downward at 2.0 m/s². What normal force does the floor apply to the person?
solution
Let up be positive, so the acceleration is negative.
$$F_g=mg$$
$$F_g=(60)(9.8)$$
$$F_g=588\,\mathrm{N}$$
$$F_N-F_g=ma$$
$$F_N-588=(60)(-2.0)$$
$$F_N=468\,\mathrm{N}$$
The normal force is smaller than the person's weight.
Example: A backpack weighs 735 N on Earth. What is its mass?
solution
$$F_g=mg$$
$$m=\frac{F_g}{g}$$
$$m=\frac{735\,\mathrm{N}}{9.8\,\mathrm{m/s^2}}$$
$$m=75\,\mathrm{kg}$$
Example: A 3000 g rock is taken to the Moon and to Mars. What is its weight on each world? Use 1.6 m/s² for the Moon and 3.711 m/s² for Mars.
solution
The mass stays the same, but weight changes because gravity changes.
$$3000\,\mathrm{g}=3.0\,\mathrm{kg}$$
$$\text{Moon}$$
$$F_g=mg$$
$$F_g=(3.0)(1.6)$$
$$F_g=4.8\,\mathrm{N}$$
$$\text{Mars}$$
$$F_g=mg$$
$$F_g=(3.0)(3.711)$$
$$F_g=11.1\,\mathrm{N}$$
Example: A sample weighs 22.3 N on Mars, where gravity is 3.711 m/s². What is the sample's mass?
solution
$$F_g=mg$$
$$m=\frac{F_g}{g}$$
$$m=\frac{22.3\,\mathrm{N}}{3.711\,\mathrm{m/s^2}}$$
$$m=6.01\,\mathrm{kg}$$
Example: A falling crate has a weight of 686 N downward and air resistance of 196 N upward. The crate is 1.2 m tall, but treat it as a single object. What is its mass and acceleration?
solution
First find the mass from the weight.
$$F_g=mg$$
$$m=\frac{686\,\mathrm{N}}{9.8\,\mathrm{m/s^2}}$$
$$m=70\,\mathrm{kg}$$
Let up be positive.
$$\sum F=196\,\mathrm{N}-686\,\mathrm{N}$$
$$\sum F=-490\,\mathrm{N}$$
$$\sum F=ma$$
$$a=\frac{\sum F}{m}$$
$$a=\frac{-490\,\mathrm{N}}{70\,\mathrm{kg}}$$
$$a=-7.0\,\mathrm{m/s^2}$$
The crate accelerates downward.
Example: A skydiver has a mass of 70 kg and is falling at terminal velocity, so the acceleration is zero. What upward air resistance force acts on the skydiver?
solution
At terminal velocity, the net force is zero.
$$F_g=mg$$
$$F_g=(70)(9.8)$$
$$F_g=686\,\mathrm{N}$$
$$F_{\mathrm{air}}-F_g=0$$
$$F_{\mathrm{air}}=686\,\mathrm{N}$$
Example: A 10 kg box slides without friction on a 30° ramp. Find the force of gravity, the normal force, the downhill component of gravity, and the acceleration.
solution
$$F_g=mg$$
$$F_g=(10)(9.8)$$
$$F_g=98\,\mathrm{N}$$
Perpendicular to the ramp:
$$F_{g\perp}=F_g\cos(30^\circ)$$
$$F_{g\perp}=(98)\cos(30^\circ)$$
$$F_{g\perp}=84.9\,\mathrm{N}$$
$$F_N=84.9\,\mathrm{N}$$
Parallel to the ramp:
$$F_{g\parallel}=F_g\sin(30^\circ)$$
$$F_{g\parallel}=(98)\sin(30^\circ)$$
$$F_{g\parallel}=49.0\,\mathrm{N}$$
$$a=\frac{49.0\,\mathrm{N}}{10\,\mathrm{kg}}$$
$$a=4.9\,\mathrm{m/s^2}$$
Example: A 25 kg sled slides without friction down a 20° hill. Find the normal force and the acceleration down the hill.
solution
$$F_g=mg$$
$$F_g=(25)(9.8)$$
$$F_g=245\,\mathrm{N}$$
$$F_{g\perp}=F_g\cos(20^\circ)$$
$$F_{g\perp}=(245)\cos(20^\circ)$$
$$F_N=230\,\mathrm{N}$$
$$F_{g\parallel}=F_g\sin(20^\circ)$$
$$F_{g\parallel}=(245)\sin(20^\circ)$$
$$F_{g\parallel}=83.8\,\mathrm{N}$$
$$a=\frac{83.8\,\mathrm{N}}{25\,\mathrm{kg}}$$
$$a=3.35\,\mathrm{m/s^2}$$
Example: A 16 kg object has 85 N downward, 22 N upward, 140 N left, 78 N right, and 55 N right acting on it. Find the horizontal and vertical acceleration.
solution
Let right and up be positive.
$$\text{horizontal}$$
$$\sum F_x=78+55-140$$
$$\sum F_x=-7\,\mathrm{N}$$
$$a_x=\frac{-7\,\mathrm{N}}{16\,\mathrm{kg}}$$
$$a_x=-0.44\,\mathrm{m/s^2}$$
$$\text{vertical}$$
$$\sum F_y=22-85$$
$$\sum F_y=-63\,\mathrm{N}$$
$$a_y=\frac{-63\,\mathrm{N}}{16\,\mathrm{kg}}$$
$$a_y=-3.94\,\mathrm{m/s^2}$$
Example: A 4 kg lamp hangs motionless from a vertical cable. What is the tension in the cable?
solution
The lamp is not accelerating, so the net force is zero.
$$F_g=mg$$
$$F_g=(4)(9.8)$$
$$F_g=39.2\,\mathrm{N}$$
$$T-F_g=0$$
$$T=39.2\,\mathrm{N}$$
Example: A 15 kg crate is lifted upward by a rope with acceleration 1.2 m/s². What is the tension in the rope?
solution
Let up be positive.
$$F_g=mg$$
$$F_g=(15)(9.8)$$
$$F_g=147\,\mathrm{N}$$
$$T-F_g=ma$$
$$T-147=(15)(1.2)$$
$$T=165\,\mathrm{N}$$
Example: A 20 kg box sits in an elevator moving upward at 3.0 m/s. What is the normal force on the box?
answer
This cannot be solved from the information given. The elevator's speed does not determine the normal force. You need the elevator's acceleration.
Example: A 12 kg crate is lowered by a rope with acceleration 0.75 m/s² downward. What is the tension in the rope?
solution
Let up be positive, so the acceleration is negative.
$$F_g=mg$$
$$F_g=(12)(9.8)$$
$$F_g=117.6\,\mathrm{N}$$
$$T-F_g=ma$$
$$T-117.6=(12)(-0.75)$$
$$T=108.6\,\mathrm{N}$$
Example: A dog pulls forward with 80 N while a leash pulls backward and upward at 45°. The dog is not accelerating. What is the total tension in the leash?
solution
The horizontal part of the tension must balance the dog's 80 N pull.
$$T_x=80\,\mathrm{N}$$
For a 45 degree rope, the horizontal and vertical parts are equal.
$$T_y=80\,\mathrm{N}$$
$$T^2=T_x^2+T_y^2$$
$$T^2=(80)^2+(80)^2$$
$$T=113\,\mathrm{N}$$
Example: A 30 kg cart pulls a 10 kg cart with a rope on a frictionless floor. A 0.080 kN force pulls the 30 kg cart forward. What is the acceleration of the system and the tension in the rope?
solution
Treat both carts as one system to find acceleration.
$$0.080\,\mathrm{kN}=80\,\mathrm{N}$$
$$\sum F=ma$$
$$80\,\mathrm{N}=(30\,\mathrm{kg}+10\,\mathrm{kg})a$$
$$a=2.0\,\mathrm{m/s^2}$$
Now look only at the 10 kg cart. The rope tension is the force that accelerates it.
$$T=ma$$
$$T=(10)(2.0)$$
$$T=20\,\mathrm{N}$$
Example: A 5 kg cart pulls a 3 kg cart with a rope on a frictionless floor. Both accelerate left at 2 m/s². What is the rope tension, and what applied force pulls the 5 kg cart left?
solution
Use the 3 kg cart first.
$$T=ma$$
$$T=(3)(2)$$
$$T=6\,\mathrm{N}$$
The applied force must accelerate both carts.
$$F_{\mathrm{applied}}=(5\,\mathrm{kg}+3\,\mathrm{kg})(2\,\mathrm{m/s^2})$$
$$F_{\mathrm{applied}}=16\,\mathrm{N}$$
Example: A 12 kg block on a frictionless table is connected by a massless rope over a pulley to a hanging 4 kg block. What is the acceleration and rope tension?
solution
The hanging block's weight pulls the whole system.
$$F_g=(4)(9.8)$$
$$F_g=39.2\,\mathrm{N}$$
Use both masses to find the system acceleration.
$$F_g=(12+4)a$$
$$39.2=16a$$
$$a=2.45\,\mathrm{m/s^2}$$
The table block is pulled only by tension.
$$T=(12)(2.45)$$
$$T=29.4\,\mathrm{N}$$
Example: Two hanging masses are connected over a frictionless pulley. One mass is 8 kg and the other is 5 kg. What is the acceleration and rope tension?
solution
The heavier mass moves down and the lighter mass moves up. The driving force is the difference in their weights.
$$F_{\mathrm{drive}}=(8)(9.8)-(5)(9.8)$$
$$F_{\mathrm{drive}}=29.4\,\mathrm{N}$$
$$F_{\mathrm{drive}}=(8+5)a$$
$$29.4=13a$$
$$a=2.26\,\mathrm{m/s^2}$$
Use the 5 kg mass to find tension.
$$T-F_g=ma$$
$$T-(5)(9.8)=(5)(2.26)$$
$$T=60.3\,\mathrm{N}$$
Example: Two hanging masses are connected over a frictionless pulley. One mass is 15 kg and the other is 9 kg. What is the acceleration and rope tension?
solution
$$F_{\mathrm{drive}}=(15)(9.8)-(9)(9.8)$$
$$F_{\mathrm{drive}}=58.8\,\mathrm{N}$$
$$58.8=(15+9)a$$
$$a=2.45\,\mathrm{m/s^2}$$
Use the lighter mass, which accelerates upward.
$$T-F_g=ma$$
$$T-(9)(9.8)=(9)(2.45)$$
$$T=110\,\mathrm{N}$$
Example: An 18 kg box is held at rest on a frictionless 25° ramp by a rope pulling up the ramp. What are the tension and normal force?
solution
First find the weight.
$$F_g=mg$$
$$F_g=(18)(9.8)$$
$$F_g=176.4\,\mathrm{N}$$
The rope must balance the downhill part of gravity.
$$T=F_g\sin(25^\circ)$$
$$T=(176.4)\sin(25^\circ)$$
$$T=74.6\,\mathrm{N}$$
The normal force balances the perpendicular part of gravity.
$$F_N=F_g\cos(25^\circ)$$
$$F_N=(176.4)\cos(25^\circ)$$
$$F_N=160\,\mathrm{N}$$
Example: A 70 kg person stands on a scale in an elevator. The scale reads 0.735 kN. What is the elevator's acceleration?
solution
The scale reading is the normal force.
$$0.735\,\mathrm{kN}=735\,\mathrm{N}$$
$$F_g=mg$$
$$F_g=(70)(9.8)$$
$$F_g=686\,\mathrm{N}$$
Let up be positive.
$$F_N-F_g=ma$$
$$735-686=(70)a$$
$$a=0.70\,\mathrm{m/s^2}$$
The elevator accelerates upward.
Example: A 100 kg package falls with 260 N of air resistance upward. What is its acceleration?
solution
Let up be positive.
$$F_g=mg$$
$$F_g=(100)(9.8)$$
$$F_g=980\,\mathrm{N}$$
$$\sum F=260\,\mathrm{N}-980\,\mathrm{N}$$
$$\sum F=-720\,\mathrm{N}$$
$$a=\frac{-720\,\mathrm{N}}{100\,\mathrm{kg}}$$
$$a=-7.2\,\mathrm{m/s^2}$$
The package accelerates downward.
Example: A 5000 g sign hangs at rest from two vertical ropes. The ropes share the force equally. What is the tension in each rope?
solution
$$5000\,\mathrm{g}=5\,\mathrm{kg}$$
Find the sign's weight.
$$F_g=mg$$
$$F_g=(5)(9.8)$$
$$F_g=49\,\mathrm{N}$$
The two ropes share the upward force equally.
$$2T=49\,\mathrm{N}$$
$$T=24.5\,\mathrm{N}$$
Example: A 20 kg sled is pulled across frictionless ice by a rope with 100 N of tension at 30° above horizontal. What are the normal force and horizontal acceleration?
solution
Break the tension into parts.
$$T_x=T\cos(30^\circ)$$
$$T_x=(100)\cos(30^\circ)$$
$$T_x=86.6\,\mathrm{N}$$
$$T_y=T\sin(30^\circ)$$
$$T_y=(100)\sin(30^\circ)$$
$$T_y=50.0\,\mathrm{N}$$
Vertical acceleration is zero.
$$F_N+T_y-F_g=0$$
$$F_N+50.0-(20)(9.8)=0$$
$$F_N=146\,\mathrm{N}$$
Horizontal motion:
$$a_x=\frac{T_x}{m}$$
$$a_x=\frac{86.6\,\mathrm{N}}{20\,\mathrm{kg}}$$
$$a_x=4.33\,\mathrm{m/s^2}$$
Example: A 30 kg cart starts from rest on a frictionless floor. A rope pulls it horizontally with 90 N for 0.0667 min. What is the cart's acceleration, and how far does it move?
solution
First convert the time, then use force to find acceleration.
$$0.0667\,\mathrm{min}=4.0\,\mathrm{s}$$
$$\sum F=ma$$
$$90\,\mathrm{N}=(30\,\mathrm{kg})a$$
$$a=3.0\,\mathrm{m/s^2}$$
Now use acceleration in a motion equation.
$$\Delta x=u\Delta t+\tfrac{1}{2}a\Delta t^2$$
$$\Delta x=(0)(4.0)+\tfrac{1}{2}(3.0)(4.0)^2$$
$$\Delta x=24\,\mathrm{m}$$
Example: A 1200 kg car moving at 72 km/h must stop in 50 m. What constant net force is needed to stop it?
solution
First convert the speed, then use motion to find the acceleration. Let forward be positive.
$$72\,\mathrm{km/h}=20\,\mathrm{m/s}$$
$$v^2=u^2+2a\Delta x$$
$$(0)^2=(20)^2+2a(50)$$
$$0=400+100a$$
$$a=-4.0\,\mathrm{m/s^2}$$
Now use Newton's second law.
$$\sum F=ma$$
$$\sum F=(1200\,\mathrm{kg})(-4.0\,\mathrm{m/s^2})$$
$$\sum F=-4800\,\mathrm{N}$$
The force must be 4800 N opposite the car's motion.
Example: A 10 kg block on a frictionless 30° ramp is connected over a light pulley to a hanging 6 kg block. The ramp is 2.0 m long, but that length is not needed for the force calculation. Find the acceleration and the tension. The 6 kg block moves downward.
solution
For the block on the ramp, the downhill component of gravity is:
$$F_{g\parallel}=mg\sin(30^\circ)$$
$$F_{g\parallel}=(10)(9.8)\sin(30^\circ)$$
$$F_{g\parallel}=49.0\,\mathrm{N}$$
The hanging block's weight is:
$$F_g=(6)(9.8)$$
$$F_g=58.8\,\mathrm{N}$$
The driving force on the two-object system is the difference.
$$F_{\mathrm{drive}}=58.8\,\mathrm{N}-49.0\,\mathrm{N}$$
$$F_{\mathrm{drive}}=9.8\,\mathrm{N}$$
$$F_{\mathrm{drive}}=(10+6)a$$
$$9.8=16a$$
$$a=0.61\,\mathrm{m/s^2}$$
Use the block on the ramp to find tension.
$$T-F_{g\parallel}=ma$$
$$T-49.0=(10)(0.61)$$
$$T=55.1\,\mathrm{N}$$
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How does its mass and weight change in a freefall on Earth?
Example: The hardcover version of Seveneves by Neal Stephenson has a mass of 0.95 kg.
What is the force of gravity felt by the book on Earth? What about on the Moon?
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