/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}

91影视

In Figure 11.26, if rA=3m, and =30, what is the magnitude of the torque about locationAincluding units? If the force in Figure 11.26 were perpendicular to rA but gave the same torque as before, what would be its magnitude?

Short Answer

Expert verified

The magnitude of torque at location A is6N.m.

The magnitude of the force relative to location A is2N .

Step by step solution

01

Definition of Torque

The force that can cause an object to twist along an axis is measured as torque. In linear kinematics, force is what causes an object to accelerate. Torque is also responsible for the angular acceleration. As a result, torque can be defined as the linear force's rotational equivalent.

02

The given data

Given:

Force acting on the ranch,F=4N

Force is at an angle,=30

Length of the ranch,rA=3m

The figure given below shows the direction of the force on the wrench.

03

Find the magnitude of the force relative to location A 

Magnitude of the torque relative to location A is given by formula:

Torque, =rAFsin

Substitute the values in the above equation

=(3m)(4N)sin30=6N.m

Hence, the magnitude of torque at location A is 6N.m.

To produce the same torque, the amount of the force acting on the torque relative to position A is given by-

F=rAsin

F=(6N.m)(3m)sin90=2N

Hence, the magnitude of force at location A is 2N.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A stationary bicycle wheel of radiusis mounted in the vertical plane on a horizontal low-friction axle (Figur The 11.43).Thewheel has mass,M all concentrated in the rim (the spokes have negligible mass). A lump of clay with mass m falls and sticks to the outer edge of the wheel at the location shown. Just before the impact the clay has a speed v(a) Just before the impact, what is the angular momentum of the combined system of wheel plus clay about the center C (b) Just after the impact, what is the angular momentum of the combined system of wheel plus clay about the centerCin terms of the angular speed of the wheel? (c) Just after the impact, what are the magnitude and direction of the angular velocity of the wheel? (d) Qualitatively, what happens to the linear momentum of the combined system? Why?

A barbell spins around a pivot at A its center at (Figure). The barbell consists of two small balls, each with mass m=0.4kgat the ends of a very low mass road of length d=0.6m. The barbell spins clockwise with angular speed0=20 radians/s. (a) Consider the two balls separately, and calculateLtrans,1,A and Ltrans,2,A(both direction and magnitude in each case). (b) Calculate Ltot,A=Ltrans,I,A+Ltrans,2,A(both direction and magnitude). (c) Next, consider the two balls together and calculate for the barbell. (d) What is the direction of the angular velocity 0? (e) Calculate Lrot=I0 (both direction and magnitude). (f) How does Lrotcompare to Lrot,A? The point is the form I is just a convenient way of calculating the (rotational) angular momentum of multiparticle system. In principle one can always calculate the angular momentum simply by adding up the individual angular momentum of all the particles. (g) calculate Krot.

Design a decorative 鈥渕obile鈥 to consist of a low-mass rod of length 0.49msuspended from a string so that the rod is horizontal, with two balls hanging from the ends of the rod. At the left end of the rod hangs a ball with mass 0.484kg.At the right end of the rod hangs a ball with mass 0.273kg.You need to decide how far from the left end of the rod you should attach the string that will hold up the mobile, so that the mobile hangs motionless with the rod horizontal (鈥渆quilibrium鈥). You also need to determine the tension in the string supporting the mobile. (a) What is the tension in the string that supports the mobile? (b) How far from the left end of the rod should you attach the support string?

At t=15s, a particle has angular momentum3,5,-2kgm2/s relative to locationA . A constant torque10,-12,20Nm relative to locationA acts on the particle. Att=15.1s. what is the angular momentum of the particle?

A solid wood top spins at high speed on the floor, with a spin direction shown in figure 11.112

a. Using appropriately labeled diagrams, explain the direction of motion of the top (you do not need to explain the magnitude).

b. How would the motion change if the top had a higher spin rate? Explain briefly.

c. If the top were made of solid steel instead of wood, explain how this would affect the motion (for the same spin rate).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.