## Newton’s laws and weightlessness

Source: “Physics for Scientist and Engineers” Serway and Jewett, 6th edition

#### 6.25 A person stands on a scale in an elevator. As the elevator starts, the scale has a constant reading of 591 N. As the elevator later stops, the scale reading is 391 N. Assume the magnitude of the acceleration is the same during starting and stopping, and determine (a) the weight of the person, (b) the person’s mass, and (c) the acceleration of the elevator.

Analyze the problem:

The problems mentions two stages of motion: (1) starting of motion and (2) stopping of motion for an elevator. The given forces are the normal forces or the readings of the scale.

What are given information:

1. Normal force (N1) when elevator is starting = 591 N
2. Normal force (N2) when elevator is stopping = 391 N

What the problem need:

1. Weight of person
2. Person’s mass
3. Acceleration of the elevator

SOLUTION:

Now, we have to recall Newton’s laws to be able to analyze the motion of man in the elevator.

By the starting of motion an acceleration of the elevator will be directed toward a specific direction and when it stops it will change to the other opposite direction.

## Newton’s Laws and Circular Motion

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

6.11 A 4.00-kg object is attached to a vertical rod by two strings, as in Figure bellow. The object rotates in a horizontal circle at constant speed 6.00 m/s. Find the tension in (a) the upper string and (b) the lower string.

Analyze the problem:

The problems states that it wants tension force in string for the above setting. Newton’s laws should be used beside the dynamic of circular motion concepts.

What are given information:

1. Strings lengths (L) = 2 m
2. Height (h) = 3 m
3. Mass of object (m) = 4 kg
4. Speed of object (v) = 6 m/s

What the problem need:

1. Tension of upper string
2. Tension of lower string

SOLUTION:

From above figure we can see that we have isosceles triangle hence we can have right triangle by drawing a perpendicular line on the base of triangle to get sides lengths 2 m, and 3/2=1.5 m, so:

Then:

The tension in the upper string and  the lower string

## Simple Dynamics of Circular Motion Problem

From Serway and Jewett Physics for Scientists and Engineers with modern physics, 9th edition

Let’s analyze the wording of the problem:

1. A conical pendulum is a swinging pendulum that can rotate around an axis. The bob (the mass) will have circular path at all angles θ, but here we need to investigate when the angle is 5°.
2. When a body moves is circular motion it must have at least a radial acceleration (centripetal acceleration).
3. The mass is connected by a wire, the wire will have a tension (because it is used to suspend the bob). The tension should have x- and y-components.

What the information give in problem:

1. The bob’s mass (m) = 80 kg
2. Length of wire (L) = 10 m
3. Angle of pendulum (θ) = 5°

What the problem need:

1. The x- and y-components of tension force.

SOLUTION:

• The x- and y-components of tension force.
• The x- and y-components of the tension force are:

## Newton’s Laws of Motion

From Young and Freedman’s University Physics with Modern Physics textbook, 14th edition

Let’s analyze the wording of the problem:

1. An astronaut inside a rocket that moves vertically upward “An astronaut is inside…rocket…blasting off vertically”. The motion is in 1 dimension (y-axis).
2. The speed the rocket needs to reach is 331 m/s and as quick as possible but reaching that speed in an acceleration above 4g could make the astronauts blackout.
3. by reading the word thrust it will definitely mean using Newton’s laws of motion.

What are the given information:

1. Rocket mass (M) = 2.25×10^6
2. Initial velocity of rocket (v0)=0 m/s
3. Final velocity of rocket (vf)=331 m/s
4. Maximum possible acceleration of rocket = 4g
5. Maximum possible acceleration of the rocket will give us the quickest blast to 331 m/s without astronauts blackout, then it is constant through out the motion of rocket until it reaches 331 m/s.

The problem needs:

1. Maximum initial thrust (the force that takes the rocket up).
2. The force the rocket exert on astronaut (in terms of astronaut’s weight).
3. The shortest time the rocket can reach the speed of sound.

THE SOLUTION:

1. Maximum initial thrust (the force that takes the rocket up).

2.The force the rocket exert on astronaut (in terms of astronaut’s weight).

3. The shortest time the rocket can reach the speed of sound.

## Newton’s Laws Problem

From Young and Freedman’s University Physics with Modern Physics Textbook, 14th edition

Let’s analyze the wording of the problem:

1. “…bullet…strikes a large tree…” and “…assume a constant retarding force…”, gives hints that Newton’s laws would be involved.
2. “What force… the tree exert on the bullet?” means Newton’s 2nd and 3rd laws will be used.

What information given in this problem:

1. Speed of bullet = 350 m/s
2. Penetration depth = 0.13 m = distance traveled
3. Bullet mass = 1.8 g
4. 0.22-caliber rifle not an important information for solving the problem.

What the problem need:

1. The time required for the bullet to stop inside the tree.
2. The force the tree exert on the bullet.

SOLUTION:

## Simple Newton’s laws problem

From Young and Freedman’s University Physics with modern physics textbook, 14th edition

Find the hints of type of problem:

1.  “… producing the force vectors…” the two words “force vector” hints using Newton’s law.
2. “…find the magnitude and direction of the resultant…” then Newton’s 2nd law would be used.

The given information:

1. Force vector 1 magnitude = 985 N and direction 31° with the positive x-axis.
2. Force vector 2 magnitude =788 N and direction 32° with the positive y-axis (it means 90°+32° = 122° from positive x-axis counter clockwise).
3. Force vector 3 magnitude = 411 N and direction 53° with the negative x-axis (it means 180°+53° = 233° from positive x-axis counter clockwise).

What the problem need:

1. The x- and y-axis components of each force vector.
2. The resultant force vector (magnitude and direction).

SOLUTION:

1. The x- and y-axis components of each force vector are:

2. The resultant force vector (magnitude and direction)

From looking at the vector of the resultant force equation, the vector is at the first quarter and that makes the angle in the first quarter from the positive x-axis counter clockwise.

## Newton’s Laws Problem

From past exam (Kuwait University), Summer 2016 2nd midterm.

The wording of the problem:

1. Objects with masses that are in physical contact (…Two small blocks…connected on a large block…”, a hint that it might be a problem involves Newton’s laws.
2. “…Horizontal force F…”, means Newton’s laws would be used.
3. “…small blocks remain stationary relative to M?”, Would mean that Newton’s 3rd law would be used.

The given information:

1. Small blocks masses (m1 and m2) = 1 kg
2. The large block mass (M) = 10 kg
3. m1 and m2 connected by massless robe through a massless and frictionless pulley.
4. Block M on wheels and its surface is frictionless (m1 and m2 not affected by frictional force).

The needed information:

1. What could be the force F (shown above) if we do not want m1 and m2 to move at all?

SOLUTION:

The forces of interest shown in the figure below: