Explanation Given that the opposite side of the beam begins to rise when the magnitude of the force reaches 2.0 × 10 2 N . The arrangement is given in Figure 14P.29. The forces acting on the beam is shown in Figure 1. There are two forces, the gravitational force acting at the center of mass of the beam and the applied force acting at the end of the beam in the system. For equilibrium condition of the beam, the net torque has to be zero. Write the expression for the force of gravity acting on the beam. F → g = − F g j ^ (I) Here, F → g is the force of gravity, j ^ is the unit vector along y direction. Write the expression for the applied force (by pushing) on the beam. F → push = − F j ^ (II) Here, F → push is the applied force. Write the expression for the torque due to the force of gravity on the beam. τ → 1 = r → 1 × F → g (III) Here, τ → 1 is the torque due to the force of gravity, and r → 1 is the position vector of point at which the force acting from the center of rotation. Write the expression for the torque due to the force of gravity on the beam. τ → 1 = r → 1 × F → g (IV) Here, τ → 1 is the torque due to the force of gravity, and r → 1 is the position vector of point at which the force F → g acting from the center of rotation. Write the expression for the torque due to the applied force on the beam. τ → 2 = r → 2 × F → push (V) Here, τ → 2 is the torque due to the applied force, and r → 2 is the position vector of point at which the force F → push acting from the center of rotation

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ISBN: 8220100546716

Author: Katz

Publisher: CENGAGE L

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Chapter 14, Problem 29PQ

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The weight of the beam.

Solution Summary: The author illustrates the arrangement of the forces acting on the beam in Figure 14P.29.

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The weight of the beam.

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Chapter 14 Solutions