PH212 Lecture 2: Chapter-8-Rotational-Equilibrium-and-Dynamics

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olivedeer219

25 Dec 2018

School

Benjamin Franklin Institute of Technology

Department

Physics

Course

PH212

Professor

Jones

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Document Summary

With the forces acting where they are, not at the center of mass (see below) Each can be analyzed separately and then combined. Ability of force to rotate object around some axis is measured by quantity called torque (symbol t, tau) Torque depends upon magnitude and direction of force f and on position where f is applied. The perpendicular distance from the axis of rotation to a line drawn along direction of f (dsin ) What is the equation for torque? t = f [d sin ] F1 is 15 n upward 14 cm from axis. F2 is 11 n downward 7. 0 cm from axis. tnet? tnet = f1 d1 + f2 d2. What is the second condition for equilibrium? tnet = ti = 0 necessary for rotational equilibrium. It"s arbitrary; however, a clever choice can simplify problem best place is to choose one through which an unknown force passes so its torque contribution will be zero.

Related Questions

A force F⃗ of magnitude F making an angle θ with the x-axis is applied to a particle located along the axis of rotation A, at Cartesian coordinates (0,0) in the figure. The vector F⃗ lies in the XY plane, and the four axes of rotation A, B, C, and D all lie perpendicular to the XY plane. (Figure 1)

A particle is located at a vector position r⃗ with respect to an axis of rotation (thus r⃗ points from the axis to the point at which the particle is located). The magnitude of the torque τ about this axis due to a force F⃗ acting on the particle is given by

τ=rFsin(α)=(moment arm)⋅F=rF⊥,

where α is the angle between r⃗ and F⃗ , r is the magnitude of r⃗ , F is the magnitude of F⃗ , the component of r⃗ that is perpendicular to F⃗ is the moment arm, and F⊥ is the component of the force that is perpendicular to r⃗ .

Sign convention: You will need to determine the sign by analyzing the direction of the rotation that the torque would tend to produce. Recall that negative torque about an axis corresponds to clockwise rotation.

In this problem, you must express the angle α in the above equation in terms of θ, ϕ, and/or π when entering your answers. Keep in mind that π=180degrees and (π/2)=90 degrees.

Part A

What is the torque τA about axis A due to the force F⃗ ?

Express the torque about axis A at Cartesian coordinates (0,0).

τA =

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Part B

What is the torque τB about axis B due to the force F⃗ ? (B is the point at Cartesian coordinates (0,b), located a distance b from the origin along the y axis.)

Express the torque about axis B in terms of F, θ, ϕ, π, and/or other given coordinate data.

τB =

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Part C

What is the torque τC about axis C due to F⃗ ? (C is the point at Cartesian coordinates (c,0), a distance c along the x axis.)

Express the torque about axis C in terms of F, θ, ϕ, π, and/or other given coordinate data.

τC =

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Part D

What is the torque τD about axis D due to F⃗ ? (D is the point located at a distance d from the origin and making an angle ϕ with the x axis.)

Express the torque about axis D in terms of F, θ, ϕ, π, and/or other given coordinate data.

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The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.

(a) What is , the magnitude of the tangential force that acts on the pole due to the tension in the rope?

Express your answer in terms of

and

.

When using the tangential force method, you calculate the torque using the equation

$\tau = f_{\rm t} d$ ,

where

is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque).

(b) What is the magnitude of the torque

on the pole, about point A, due to the tension in the rope?

Express your answer in terms of

, and

The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm.

For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause.

, the length of the moment arm of the force.

Express your answer in terms of

and

(d)Now consider a woman standing on the ball of her foot as shown. A normal force of magnitude

acts upward on the ball of her foot. Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a force

. This small bone has a length

, and the angle between this bone and the Achilles' tendon is

. The horizontal displacement between the ball of the foot and the point P is

Find τp, the torque about point P due to the force applied by the Achilles' tendon. Express your answer in terms of F, ϕ, and x.

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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

PH212 Lecture 2: Chapter-8-Rotational-Equilibrium-and-Dynamics

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