8.01 Lecture Notes - Lecture 1: Friction, Mass, Free Body

67 views2 pages

amethystmuskrat145

13 Apr 2016

School

Department

Course

Professor

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Everybody continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it. The change of motion is proportional to the motive force impresses, and is made in the direction of the right line in which that force is impressed. To every action there is always opposed an equal reaction: or, the mutual action of two bodies upon each other are always equal, and directed to contrary parts. Frictional forces: static and kinetic friction, uid resistance. Gravitational force near the surface of the earth. A rope is attached to a block b on one end, and pulled by an applied force from the other end. The force of the rope f on the block is called a tension force. When the rope is virtually massless the magnitude of the tension force is equal to the magnitude of the applied pulling force.

Related Questions

which of the vectors below best represents the direction of the impulse vector j⃗ ?

To learn about the impulse-momentum theorem and its applications in some common cases.

Using the concept of momentum, Newton's second law can be rewritten as

?F? =dp? dt, (1)

where ?F? is the net force F? net acting on the object, and dp? dt is the rate at which the object's momentum is changing.

If the object is observed during an interval of time between times t1 and t2, then integration of both sides of equation (1) gives

?t2t1?F? dt=?t2t1dp? dtdt. (2)

The right side of equation (2) is simply the change in the object's momentum p2??p1?. The left side is called the impulse of the net force and is denoted by J? . Then equation (2) can be rewritten as

J? =p2??p1?.

This equation is known as the impulse-momentum theorem. It states that the change in an object's momentum is equal to the impulse of the net force acting on the object. In the case of a constant net force F? net acting along the direction of motion, the impulse-momentum theorem can be written as

F(t2?t1)=mv2?mv1. (3)

Here F, v1, and v2 are the components of the corresponding vector quantities along the chosen coordinate axis. If the motion in question is two-dimensional, it is often useful to apply equation (3) to the x and y components of motion separately.

( The following questions will help you learn to apply the impulse-momentum theorem to the cases of constant and varying force acting along the direction of motion. First, let us consider a particle of mass m moving along the x-axis. The net force F is acting on the particle along the x-axis. F is a constant force.)

A) The particle starts from rest at t=0. What is the magnitude p of the momentum of the particle at time t? Assume that t>0.

B) The particle starts from rest at t=0. What is the magnitude v of the velocity of the particle at time t? Assume that t>0.

C) The particle has the momentum of magnitude p1 at a certain instant. What is p2, the magnitude of its momentum ?t seconds later?

D) The particle has the momentum of magnitude p1 at a certain instant. What is v2, the magnitude of its velocity ?t seconds later?

E) Find the magnitude of the impulse J delivered to the particle.

F) Which of the vectors below best represents the direction of the impulse vector J??
(Figure 1)

1
2
3
4
5
6
7
8

puceworm982

A Simple Atwood Machine consists of two masses m1 and m1 that are connected by a string wound over a pulley, as seen in the figure below. m2 is larger than m1. On a piece of paper, draw two free body diagrams; one for each of the masses, showing all forces acting on each mass. Then answer the following questions. (a) Based on your free body diagram, identify the forces acting on the mass m1. Select all that apply. (m1+m2)a m1g FN m1a (m1+m2)g m2a f m2g T (b) Using the direction rosette indicate the direction for each of the forces acting on m1 that you identified in part (a). If any of the forces listed below is not applicable, enter NA. m1a= m2a= m1g= m2g= (m1+m2)a= (m1+m2)g= f= T= FN= (c) Based on your free body diagram, identify the forces acting on the mass m2. Select all that apply. T (m1+m2)g m2g m1g (m1+m2)a m1a m2a f FN (d) Using the direction rosette indicate the direction for each of the forces acting on m2 that you identified in part (c). If any of the forces listed below is not applicable, enter NA. m1a= m2a= m1g= m2g= (m1+m2)a= (m1+m2)g= f= T= FN= (e) Write Newton's Second Law equation (in the form Fnet = ma) for m1 in the y-direction. Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative. (Use m1, g, a, and T as appropriate.) LHS = RHS = (f) Write Newton's Second Law equation (in the form Fnet = ma) for m2 in the y-direction. Use the first answer box for the left hand side (LHS) of the equation and the second answer box for the right hand side (RHS) of the equation. It is important to use the correct signs for each of the forces. Motion in the upward direction is positive and motion in the downward direction is negative.(Use m2, g, a, and T as appropriate.) LHS = RHS = (g) Use your answers from parts (e) and (f) to get an expression for the magnitude of the tension T. (Use m1, m2, and g as appropriate.) magnitude of T = (h) Use your answers from parts (e), (f), and (g) to find the magnitude of a, the acceleration of the system. (Use m1, m2, and g as appropriate.) magnitude of a = (i) In what direction will each mass accelerate. For mass m1: For mass m2: (j) Suppose m1 = 7 kg and m2 = 11 kg. What is the tension in the string? T = (k) Suppose m1 = 7 kg and m2 = 11 kg. What is the value of the acceleration? a = (l) Suppose that m2 starts from rest at a height of 5 m. Use the Kinematic Equations to determine how long it takes for m2 to hit the ground.

8.01 Lecture Notes - Lecture 1: Friction, Mass, Free Body

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

8.01 Lecture Notes - Lecture 1: Headon, Superposition Principle, Vector Decomposition

8.01 Lecture Notes - Lecture 1: Free Body Diagram, Polar Coordinate System, Angular Acceleration

EXCI 351 Chapter Notes - Chapter 4: Net Force, Ground Reaction Force, Normal Force

Related Questions