Chapter 4 - Forces & Newton's Laws

Question 1

Newton’s First Law

Answer 1

Object continues in state of rest or motion at a constant velocity unless compelled to change that state by a net force

Vector sum of all forces 0, all forces on object balance so no change in motion

Question 2

Inertia

Answer 2

Natural tendency of an object to remain at rest or in motion at constant velocity

mass is a measure of intertia, greater

Question 3

Newton’s 2nd Law

Answer 3

F = ma

Can separate into x and y components

Question 4

Newton’s 3rd Law

Answer 4

When one object exerts force on second object exerts oppositely directed force of equal magnitude on first object

F(AB) = -F(BA)

Question 5

Gravitational Force

Answer 5

Fg = G(m1*m2)/r^2

G = 6.674 x 10^-11 (univeral gravitational constant)

Question 6

Weight

Answer 6

W = G(mE * m)/r^2

buuuuut
GmE/r^2 is equal to little g aka 9.80

so then Weight

= gm (on earth

Question 7

Equilibrium

Answer 7

Zero acceleration

Thus vector sum of net force (due to first law) is equal to 0

Set teh forces in the x and y directions to 0 and solve

Question 8

Two forces F1 and F2 are applied to an object whose mass is 8.0 kg. The larger force is F1. When both forces point due east, the object’s acceleration has a magnitude of 0.50 m/s2. However, when F1 points due east and F2 points due west, the acceleration is 0.40 m/s2, due east. Find (a) the magnitude of F1 and (b) the magnitude of F2.

Answer 8

We have two variables, hence we need two equations. Since only the magnitude matters:

1. F1 + F2 = 8 * 0.5 = 4N
2. F1 - F2 = 8 * 0.4 = 3.2 N

Now rearrange and input into the other equation and solve

Ans: F1 = 3.6N and F2 = 0.4N

Question 9

How can you calculate the total force of perpendicular forces?

Answer 9

Use pythagorean theorem for the x component and y component forces

Question 10

Formula for apparent weight

Answer 10

Fn = mg + ma
(actual weight + factoring in acceleration)

In N

Question 11

A person pushes on a 57-kg refrigerator with a horizontal force of −267 N; the minus sign indicates that the force points in the −x direction. The coefficient of static friction is 0.65. (a) If the refrigerator does not move, what are the magnitude and direction of the static frictional force that the floor exerts on the refrigerator?

Answer 11

267N but in positive direction.

Net force acting on fridge is equal to 0 so frictional force must counteract the applied force to equal 0

Question 12

How to calculate tension with problems

Answer 12

If tension is upwards, mg opposes tension (weight).

Question 13

A light spring having a spring constant of k = 125 N/m is used to pull a 9.50 kg sled on a horizontal ice rink. The coefficient of kinetic friction
between the sled and the ice is μk = 0.200. If the sled has an accelerationof a = 2.00 m/s 2 , b y how much does the spring stretch if it pulls on the
sled:
(a) horizontally, or
(b) at 30.0° above the horizontal?

Answer 13

A) Horizontally

Draw free body diagram

Fspring - Fk = Fnet

Ffriction = us x Fn (normal force is mg)
= 0.2 * 9.5 * 9.8

F net = ma
= 9.50 * 2

Then isolate and solve for F spring and isolate that for the equation
F spring = kx

B) Above horizontal
Changes dynamics a bit since normal force is now counteracting both the forces of weight AND upwards force of spring

Analyze the forces in both the x and y direction

x direction:
spring force and friction

ma = kxcos30 - ukN

y direction:
spring force, normal and weight

0 = kxsin30 + N - mg

what is important to note is that the sled itself is not on an incline, it is just being pulled on an angle. which means that there is no vertical/horizontal components of the friction nor normal force.

Now you isolate for N from the fy equation and sub into the fx equation and solve for x! you should come out to 0.311

Question 14

A block (mass M = 0.25 kg) is at rest on a rough inclined plane (kinetic
coefficient of friction μk = 0.35 and angle θ = 30o) and is connected to an
object with mass m = 0.40 kg which hangs freely, as shown. The rope
may be considered massless; and the pulley may be considered massless
and frictionless. Find the speed v of mass m when it has fallen distance
y = 1.0 m

Answer 14

Draw the free body diagram showing this quesiton. You should make note of waht exactly the normal force equals.

For M, the tension points up the incline and the friction and weight point downwards.

The coordinate system should be set up in such a way that downwards is positive, hence the tension on m should be negative

Youknow the equations F = ma, and for each block include the tension as subtraction for the net force calculations.

for m: T-mg=-ma (acceleration has to be negative if we set it equal)
T = 3.92 - 0.4a

for M: T- mgsinx - mgcosxus = ma
T - 1.96625 = 0.25a

sub 2 into 1 and solve for a
a= 3.01

use v2 =v0^2 * ay

Question 15

A person is trying to judge whether a picture (mass = 1.10 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.660. What is the minimum amount of pressing force that must be used?

Answer 15

The normal force is opposite teh pushing force so tey are equal. Friction is pointed upwards and mg pointed downwards so they are equal.

So then usFN = mg but Fn = P so then usP = mg!

Question 16

What direction does friction point in?

Answer 16

If car is accelerating, static friction helps prevent tires from slipping so it points in same direction as acceleration

Question 17

Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Answer 17

All forces need to add to 0. So to find the resultant force, use pythagorean theorem and use tan to find the angle

tan y/x

Question 18

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. Find the tension in the cable

Answer 18

Since there is no acceleration because the velocity is constnat, the tension force is equal to the friction force

Question 19

Frictional force between 70kg soccer player and the field is 308.7N. If player comes to a stop after 2 seconds, what was their inital velocity?

Answer 19

1. Use F=ma and solve for acceleration using teh frictional force and the mass of the player
2. Then sub into the kinematics equation setting vf = 0

Question 20

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 129-kg laser sensor that measures the thickness of the ice (see the drawing). The helicopter and the sensor both move only in the horizontal direction and have a horizontal acceleration of magnitude 2.84 m/s2. Ignoring air resistance, find the tension in the cable towing the sensor.

Answer 20

There is tension in both the horizontal and vertical direction

In the horizontal
Tx = ma

In the vertical

Ty = W

Calculate both of these and use pythagorean theorem to find T

Question 21

A penguin slides at a constant velocity of 5.25 m/s down an icy incline.
The incline slopes above the horizontal at an angle of 15.0°. At the
bottom of the incline, the penguin slides onto a horizontal patch of ice.
The coefficient of kinetic friction between the penguin and the ice is the
same for the incline as for the horizontal patch. How much time is
required for the penguin to slide to a halt (stop) after entering the
horizontal patch of ice?

Answer 21

For the horizontal patch, you know that

fk = ukmg
and that f = ma
setting them equal
ma = ukmg
cancelling out the mass you get
a = -ukg

You also know that

v0 = 5.25, vf = 0, a = -ukg and t = ?

so rearranging using the kinematic equation you get

t = 5.25/-ukg

to find uk now you have to consider what occurs as sliding occurs

constnat velocity means equilibrium, the horizontal forces (friction and the x component of velocity balance each other out)

Fx = mgsinx - ukmgcosx
ukmgcosx = mgsinx
uk = mgsinx/mgcosx
uk = tanx
uk = 0.2679

Sub into the time equation above and you’re golden!

Question 22

In Figure, blocks A and C have the same weight and each weighs onefourth the weight of block B. Blocks A and C are connected by a
massless cord that passes over a massless and frictionless pulley. Assume
that there is no friction between block B and the surface on which it
moves and that the coefficient of kinetic friction between blocks A and B
is µk
= 0.2. After the system is released, [Note that A and B move with
different accelerations], find:
a. the distance block A has moved in one second, and
b. the distance block B has moved in one second.

Answer 22

For this problem, consider the downward direction as positive y.

PART A

Net force on block a (vertical is balanced)
Fa = T - ukmg
ma = T - ukmg
T = ma + ukmg

Net force on block c (horizontal is balanced)
Fa = mg - T
ma = mg - T
T = mg - ma

Now set both equations equal

mg - ma = ma + ukmg

Collect the a’s on both sides
2ma = ukmg - mg
a = g(1-uk)/2
solve for a which ends up being 3.92

Plug into the equation

x = vot + 1/2at^2
you know v0 = 0 and when you plug in you get x =1.96

PART B

Only horizontal force acting on the object is friction.

So ffriction = 4mab
ukmg = 4mab
aB = ukg/4
aB = 0.49

Sub this into the kinematic equation and you should get 0.245m