PHYS201 Lecture Notes - Lecture 12: Net Force, Acceleration

52 views2 pages

greenhippopotamus619

19 Nov 2016

School

University of Delaware

Department

Physics and Astronomy

Course

PHYS201

Professor

Gogoladze Ilia

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

Document Summary

Its value depends on the choice of a rotational axis. Torque, , is the tendency of a force to rotate an object about some axis. There is no unique value of the torque on an object. Torque increases with both the force and the distance. Only the tangential component of force causes a torque. The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. The horizontal component of the force ( cos ) has no tendency to produce a rotation. If the turning tendency of the force is counterclockwise, the torque will be positive. If the turning tendency is clockwise, the torque will be negative. The force will tend to cause a counterclockwise rotation about o. The force will tend to cause a clockwise rotation about o. Forces can cause a change in translational motion.

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 =

SubmitHintsMy AnswersGive UpReview Part

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 =

SubmitHintsMy AnswersGive UpReview Part

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 =

SubmitHintsMy AnswersGive UpReview Part

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.

apricotclam488

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.

turquoiseworm20

A rectangular loop of wire has dimensions of 0.500 m by 0.300 m. The loop is pivoted at the x-axis and lies in the XY plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop.

(a) What is the direction of the magnetic force exerted on wire segment ab?

(b) What is the direction of the torque associated with this force about an axis through the origin?

(d) What is the direction of the torque associated with this force about an axis through the origin?

(e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x-axis?

(f) Can they affect the motion of the loop in any way? Explain.

(g) What is the direction of the magnetic force exerted on segment bc?

(h) What is the direction of the torque associated with this force about an axis through the origin?

(i) What is the torque on segment ad about an axis through magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? O) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x-axis? (k) Compute the magnitude of the magnetic moment of the loop. (1) What is the angle between the magnetic moment vector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (1).

scarletbeetle53

PHYS201 Lecture Notes - Lecture 12: Net Force, Acceleration

Document Summary

Get access

Related textbook solutions

Physics: Principles with Applications

Related Documents

PHYS201 Lecture Notes - Lecture 11: Acceleration, Angular Acceleration, Speed

PHYS 100 Chapter Notes - Chapter 9: Angular Velocity, Mechanical Advantage, Free Body Diagram

PHY 101 Chapter Notes - Chapter 11: Point Particle, Angular Momentum, Right-Hand Rule

Related Questions