Exercícios halliday

displacement vectors A , B and C (and denote the result of their vector sum as r ). We choose east as the ˆi direction (+x direction) and north as the ˆj direction (+y direction). We note that the angle between C and the x axis is 60°. Thus, ( ) ( ) (50 km) ˆi (30 km) ˆj (25 km) cos 60 ˆi + (25 km)sin 60 ˆj A B C = = = ° ° (a) The total displacement of the car from its initial position is represented by r = A + B + C =(62.5 km) ˆi + (51.7 km) ˆj which means that its magnitude is r = (62.5km)2 + (51.7 km)2 = 81 km. (b) The angle (counterclockwise from +x axis) is tan–1 (51.7 km/62.5 km) = 40°, which is to say that it points 40° north of east. Although the resultant r is shown in our sketch, it would be a direct line from the “tail” of A to the “head” of C . (c) The angle between the resultant and the +x axis is given by θ = tan–1(ry/rx) = tan–1 [(10 m)/( –9.0 m)] = – 48° or 132°. Since the x component of the resultant is negative and the y component is positive, characteristic of the second quadrant, we find the angle is 132° (measured counterclockwise from +x axis). 9. We write r = a + b . When not explicitly displayed, the units here are assumed to be meters. (a) The x and the y components of r are rx = ax + bx = (4.0 m) – (13 m) = –9.0 m and ry = ay + by = (3.0 m) + (7.0 m) = 10 m, respectively. Thus r = (−9.0m) ˆi + (10m) ˆj . (b) The magnitude of r is | | 2 2 ( 9.0 m)2 (10 m)2 13 m x y r = r = r + r =…

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todos os componentes que a pasta "project" irá puxar. Inserindo a pasta library: Setup --> Set Library Root (selecione a pasta "library") --> Configuration --> Store --> User Project. Trazendo componente para o ambiente Robcad: Layout --> Get Component. Movimentando componentes no ambiente: Placement Editor --> selecione o objeto. Cinemática do Dispositivo: 1°: Apagar todos os itens obsoletos. Exemplos: pontos, linhas, curvas, frames, etc. (passo-a-passo:…

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reproduced, in any form or by any means, without permission in writing from the publisher. •2–1. If u = 30° and T = 6 kN, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. y u T x 45 8 kN 7 2 Solutions 44918 1/21/09 12:01 PM Page 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they…

reproduced, in any form or by any means, without permission in writing from the publisher. •2–1. If u = 30° and T = 6 kN, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. y u T x 45 8 kN 7 2 Solutions 44918 1/21/09 12:01 PM Page 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they…

in any form or by any means, without permission in writing from the publisher. •2–1. If u = 30° and T = 6 kN, determine the magnitude of the resultant force acting on the eyebolt and its direction measured clockwise from the positive x axis. y T u x 45 8 kN 7 2 Solutions 44918 1/21/09 12:01 PM Page 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently…

................................................................................................................................. 7 Legged robots.................................................................................................................................... 7 Hexapod ............................................................................................................................................ 8…

the sphere: the tension force [pic] of the rope (acting along the rope), the force of the wall [pic] (acting horizontally away from the wall), and the force of gravity [pic] (acting downward). Since the sphere is in equilibrium they sum to zero. Let θ be the angle between the rope and the vertical. Then Newton’s second law gives vertical component : T cos θ – mg = 0 horizontal component: FN – T sin θ = 0. (a) We solve the first equation for the tension: T…

the sphere: the tension force [pic] of the rope (acting along the rope), the force of the wall [pic] (acting horizontally away from the wall), and the force of gravity [pic] (acting downward). Since the sphere is in equilibrium they sum to zero. Let θ be the angle between the rope and the vertical. Then Newton’s second law gives vertical component : T cos θ – mg = 0 horizontal component: FN – T sin θ = 0. (a) We solve the first equation for the tension: T…