A long rigid beam with uniform mass density, a mass of 115 kg, and a length of L = 3.90 m rests on two supports placed at the ends of the beam. A student with a mass of 86.0 kg stands a distance of d = 0.90 m from the left end of the beam and the location of Support 1. d.-i Support 1 Support 2 The figure below shows an extended free-body diagram for the forces acting on the beam. You might copy the drawing and add information about the locations of the forces. S1-board S2-board П student- board W board (a) Why is it correct to draw the force due to gravity on the board (W) at the mid-point of the board? Select ALL statements that are correct. ☐ Uniform mass density means that the center of mass is located at the geometric center of the board. All of the mass of the board is concentrated at the center of the board. The board behaves as if all of its mass is located at the center of mass of the board. ☐ The center of mass is always located at the geometric center of an object. (b) If you want to use torque and rotational equilibrium to determine the magnitude and direction of the force on the board due to Support 1, which location along the beam should you use as the origin, or pivot point, for calculating torque? O the left end of the beam ○ the location of the center of the student the location of the center of the beam O the right end of the beam (c) What is the magnitude and direction of the torque due to the weight of the beam about the pivot point you selected from Part (b)? magnitude of the torque is direction of the torque is positive (CCW) O negative (CW) m.N (or N·m) O the right end of the beam (c) What is the magnitude and direction of the torque due to the weight of the beam about the pivot point you selected from Part (b)? magnitude of the torque is direction of the torque is O positive (CCW) O negative (CW) m.N (or N·m) (d) What is the magnitude and direction of the torque due to the normal force from the student on the beam about the pivot point you selected from Part (b)? magnitude of the torque is m-N (or N·m) direction of the torque is O positive (CCW) O negative (CW) (e) What is the net torque due to all forces acting on the beam? net torque = m.N (f) What is the magnitude and direction of the torque due to Support 1 on acting on the beam using the pivot point you selected from Part (b)? magnitude of the torque due to Support 1 is m.N direction of this torque is O positive (CCW) O negative (CW) (g) What is the magnitude and direction of the force due to Support 1 on acting on the beam? magnitude of the force due to Support 1 is direction of this force is ○ up down N (h) What is the magnitude and direction of the force due to Support 2 on acting on the beam? NOTE: You can use torques or forces for this. magnitude of the force due to Support 2 is N direction of this force is up ○ down

A long rigid beam with uniform mass density, a mass of 115 kg, and a length of L = 3.90 m rests on two supports placed at the ends of the beam. A student with a mass of 86.0 kg stands a distance of d = 0.90 m from the left end of the beam and the location of Support 1. d.-i Support 1 Support 2 The figure below shows an extended free-body diagram for the forces acting on the beam. You might copy the drawing and add information about the locations of the forces. S1-board S2-board П student- board W board (a) Why is it correct to draw the force due to gravity on the board (W) at the mid-point of the board? Select ALL statements that are correct. ☐ Uniform mass density means that the center of mass is located at the geometric center of the board. All of the mass of the board is concentrated at the center of the board. The board behaves as if all of its mass is located at the center of mass of the board. ☐ The center of mass is always located at the geometric center of an object. (b) If you want to use torque and rotational equilibrium to determine the magnitude and direction of the force on the board due to Support 1, which location along the beam should you use as the origin, or pivot point, for calculating torque? O the left end of the beam ○ the location of the center of the student the location of the center of the beam O the right end of the beam (c) What is the magnitude and direction of the torque due to the weight of the beam about the pivot point you selected from Part (b)? magnitude of the torque is direction of the torque is positive (CCW) O negative (CW) m.N (or N·m) O the right end of the beam (c) What is the magnitude and direction of the torque due to the weight of the beam about the pivot point you selected from Part (b)? magnitude of the torque is direction of the torque is O positive (CCW) O negative (CW) m.N (or N·m) (d) What is the magnitude and direction of the torque due to the normal force from the student on the beam about the pivot point you selected from Part (b)? magnitude of the torque is m-N (or N·m) direction of the torque is O positive (CCW) O negative (CW) (e) What is the net torque due to all forces acting on the beam? net torque = m.N (f) What is the magnitude and direction of the torque due to Support 1 on acting on the beam using the pivot point you selected from Part (b)? magnitude of the torque due to Support 1 is m.N direction of this torque is O positive (CCW) O negative (CW) (g) What is the magnitude and direction of the force due to Support 1 on acting on the beam? magnitude of the force due to Support 1 is direction of this force is ○ up down N (h) What is the magnitude and direction of the force due to Support 2 on acting on the beam? NOTE: You can use torques or forces for this. magnitude of the force due to Support 2 is N direction of this force is up ○ down

Glencoe Physics: Principles and Problems, Student Edition

1st Edition

ISBN:9780078807213

Author:Paul W. Zitzewitz

Publisher:Paul W. Zitzewitz

Chapter5: Displacement And Force In Two Dimensions

Section: Chapter Questions

Problem 82A

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