Which of the following is needed for a rigid body to be in equilibrium? (A) It must be stable. (B) The net torque on it must be zero. (C)The net force acting on it must be zero. (D)The net force and the net torque acting on it must be zero.

Which of the following is needed for a rigid body to be in equilibrium? (A) It must be stable. (B) The net torque on it must be zero. (C)The net force acting on it must be zero. (D)The net force and the net torque acting on it must be zero.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter10: Fixed-axis Rotation

Section: Chapter Questions

Problem 24CQ: Can a set of forces have a net force that is zero and a net torque that is not zero?

See similar textbooks

Similar questions

In analyzing the equilibrium of a flat, rigid object, you are about to choose an axis about which you will calculate torques. Which of the following describes the choice you should make? (a) The axis should pass through the objects center of mass. (b) The axis should pass through one end of the object. (c) The axis should be either the x axis or the y axis. (d) The axis should pass through any point within the object. (e) Any axis within or outside the object can be chosen.
Find the net torque on the wheel in Figure P10.23 about the axle through O, taking a = 10.0 cm and b = 25.0 cm. Figure P10.23
Can a set of forces have a net force that is zero and a net torque that is not zero?
In analyzing the equilibrium of a flat, rigid object, one should first choose an axis about which torques can be calculate. Which of the following is true regarding the choice of axis? a) The axis must pass through the objectť's center of mass. b) The axis must pass through one end of the object. c) The axis must be either the x axis or the y axis. d) The axis must pass through a point within the object. e) Any axis within or outside the object can be chosen.
An object has a fixed axis of rotation. A torque can be exerted on it: Oonly by a force directed radially inwards. n only by a force directed radially outwards. only by a horizontal force. by any force with a component radially outwards from the axis of rotation. by any force not pointing directly toward or away from the axis of rotation. by any force directed perpendicular to and not pointing towards the axis of rotation. Previous Page Next Page Page 14 of 14
One end of a uniform meter stick is placed against a vertical wall. The other end is held by a lightweight cord that makes an angle theta with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400. Set up equations for net force and net torque and then find the maximum value theta can have if the stick is to remain at equilibrium.
Which one of the following statements describes the condition(s) for which a rigid body is in equilibrium? Each external force may be non-zero, but the sum of the forces must be zero. Each external torque may not be non-zero, but the sum of the torques must be zero. Each external torque acting on the body must be zero. The sum of the externally applied forces must be zero, and the sum of the externally applied torques must be zero. Only internal forces can be present.
Which of the following is the condition for the rotational equilibrium? The sum of the clockwise moments about any pivot is less than the sum of the anti-clockwise moments about A. that pivot The sum of torques due to all the external forces acting on the body is greater than zero (c) The sum of torques due to all the external forces acting on the body is less than zero The sum of the clockwise moments about any pivot equal the sum of the anti-clockwise moments about that pivot Which of the following is the product of the lever-arm distance and the perpendicular force to the lever-arm? A Force B Acceleration Torque Pressure
A person bending forward to lift a load "with his back" (Figure a) rather than with his knees" can be injured by large forces exerted on the muscles and vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the principal supporting force provided by the erector spinalis muscle in the back. To see the magnitude of the forces involved, and to understand why back problems are common among humans, consider the model shown in Figure b, of a person bending forward to lift a W-195-N object. The spine and upper body are represented as a uniform horizontal rod of weight W-295 N pivoted at the base of the spine. The erector spinalls muscle, attached at a point two-thirds of the way up the spine, maintains the position of the back. The angle between the spine and this muscle is 12.0° Back muscle Pivot R₂ T120 T W W₂ 0 (a) Find the tension in the back muscle. KN D (b) Find the compressional force in the spine. (Enter the magnitude.) KN
When you apply the torque equation Er = 0 to an object in equilibrium, the axis about which torques are calculated (a) must be located at a pivot. (b) must be located at the object's center of gravity. (c) should be located at the edge of the object. (d) can be located anywhere.
A beam of mass M and length L is hanging from a rope, perpendicular to a wall. The rope is attached at a point that is 75% of the length of the beam. (a) Draw all forces acting on the beam on a figure. (b) Write the equations necessary for equilibrium, in terms of the forces for this situation. (c) Use the torque equation to solve for the tension.
(a) Where should the man sit to balance the seesaw? Apply the second condition of equilibrium to the plank by setting the sum of the torques equal to zero. The first two torques are zero. Let x represent the man's distance from the pivot. The woman is at a distance L/2 from the pivot. Solve this equation for x and evaluate it. (b) Find the normal force n exerted by the pivot on the seesaw. Apply the first condition of equilibrium to the plank, solving the resulting equation for the unknown normal force, n. (c) Repeat part (a), choosing a new axis through the left end of the plank. Compute the torques using this axis, and set their sum equal to zero. Now the pivot and gravity forces on the plank result in nonzero torques. Substitute all known quantities. Solve for x substituting the normal force found in part (b).. pivot + gravity+man+woman-0 0+ 0 Mgx + mg(L/2) - 0 m (L/2) M x 1.878 = (57.8 kg) (2.46 m) 75.7 kg m -Mg-mg - mp9 + n = 0 n = (M + m + mpig = (75.7 kg + 57.8 kg + 12.0 kg)…