Two Blocks on Inclined plane

“Physics Principles with Applications”, Douglas C. Giancoli, 7th Edition, Global Edition

44. Two blocks are connected by a light string passing over a pulley of radius 0.15 m and moment of inertia I. The blocks move (toward the right) with an acceleration of 1.00 m/s2 along their frictionless inclines (See figure below). (a) Draw free-body diagrams for each of the two blocks and the pulley. (b) Determine FTA and FTB, the tensions in the two parts of the string. (c) Find the net torque acting on the pulley, and determine its moment of inertia, I.

Giancoli 8-44

Analyze the problem:

The problem has two blocks on inclined surfaces and they are attached by an inelastic string (so they should have the same acceleration) the pulley’s mass is not ignored (we know that because the problem mentioned that there is a moment of inertia).

What are the given information?

  1. Block A mass, mA=8 kg
  2. Block B mass, mB=10 kg
  3. Acceleration, a = 1 m/s2
  4. Inclination Angle 1, θ1=32º
  5. Inclination Angle 2, θ2=61º
  6. Radius of pulley, R= 0.15 m

What are the needed information?

  1. Body-diagram draw for the two blocks and the pulley.
  2. FTA and FTB tension forces
  3. The net torque on the pulley
  4. The moment of inertia of the pulley

SOLUTION:

Giancoli 8-44-1

Giancoli 8-44-2

Atwood Machine

Source:
“Physics Principles with Applications”, Douglas C. Giancoli, 7th Edition, Global Edition

This is a common physics problem and it mentioned in several textbooks.

45. An Atwood machine (shown in figure below) consists of two masses, mA=65 kg and mB=75 kg, connected by a massless inelastic cord that passes over a pulley free to rotate, The pulley is a solid cylinder of radius R=0.45 m and mass 6 kg. (a) Determine the acceleration of each mass. (b) What % error would be made if the moment of inertia of the pulley is ignored? [Hint: the tensions FTA and FTB aare not equal]

Analyze the problem:

It is an Atwood typical problem but here the pulley’s mass  is not ignored and the tensions forces are different for the same cord. Having the pulley’s mass ignored and the tension being the same for the single cord simplify the problem, so here we have to approach the problem in different way but using the same methods (Newton’s laws!).

What are the given information?

  1. Block A mass, mA = 65 Kg
  2. Block B mass, mB=75 Kg
  3. Pulley mass, M=6 kg
  4. Pulley Radius, R=0.45 m

What are needed information?

  1. Acceleration of both blocks.
  2. Percentage error of the two cases (the pulley’s moment of inertia is ignored and the pulley’s moment of inertia is not ignored).

 

SOLUTION:

  1. Finding the acceleration of the two blocks:

Giancoli 8-45

2. Finding the percentage error:

Giancoli 8-45-2

 

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

 

 

 

 

Simple Dynamics of Circular Motion Problem

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

Serway 6.8 problem.jpg

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.
  2. The radial acceleration

 

SOLUTION:

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

serway-6-8-problem-sol-aa

  • The radial acceleration

Serway 6.8 problem -sol b.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

 

 

Simple Newton’s laws problem

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

problem-4-2

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:

problem-4-2-sol1

2. The resultant force vector (magnitude and direction)

problem-4-2-sol2

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.

summer-2016-2nd-midterm-sp10-1

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:

summer-2016-2nd-midterm-sp10-sol1

 

summer-2016-2nd-midterm-sp10-sol2