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.

6-25c.jpg

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.

Serway 6.11 problem.jpg

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:

serway-6-11-problem-sol-aFrom 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:

Serway 6.11 problem main sol.jpg

Then:

The tension in the upper string and  the lower string

Serway 6.11 problem main sol2.jpg

 

 

 

 

Newton’s Laws Problem

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

problem-4-28

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:

problem-4-28-sol1