phy207_hw7_solution

phy207_hw7_solution - Chapter 12, Number 2 The conditions...

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Chapter 12, Number 2 The conditions for equilibrium come from Newton's Second Law and its rotational analog. We need the summation of forces to be equal to zero and the summation of torques to be equal to zero for a static system. The force equation gives us two scalar equations: one for the x components and one for the y components. We use the standard coordinate representation with the positive y direction up and the positive x direction to the right. It follows that: F x.net F x R x 0 F y.net F y R y F g 0 To use the torque equation we find the torque due to each force individually and then sum them. (Note: as the book indicates, we will consider counterclockwise rotations as positive torques). Using the definition r F we can find the magnitude of each torque: r F r F sin where is the angle between r and F . Then the toruqe due to F x is Fx F x lsin because it forces a clockwise rotation while the torque due to F y is given by Fy F y l cos since the angle between F y and the radius vector is 2 . Now because the radius of the force R is zero, we have Rx Ry 0 . Now the last torque to consider is that due to the graviational force. The angle between this force and a radius is 2 so sin cos and the torque becomes Fg F g l 2 cos . The final equation is: net Fx Fy Fg F x lsin F y l cos F g l 2 cos 0 So our conditions for equilibrium are: F x.net F x R x 0 F y.net F y R y F g 0 net Fx Fy Fg F x lsin F y l cos F g l 2 cos 0 Chapter 12, Number 3 We want to find the torque about P when the support at O is not needed. (Think about this like a scale. We would like to move mass 2 until the system is balanced. )
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Note that in this drawing, all forces are in the y direction and the radii are all in the x direction. Thus, the magnitude of each torque is given by the magnitude of the force mulitplied by the radius. We call counterclockwise rotations positive and note that the normal at P exerts no torque. It follows that: n O l 2 d m 1 g l 2 d m b g d m 2 g x 0 If we set n O 0 we find: m 1 g l 2 d m b g d m 2 g x 0 . Hence we arrive at the answer x m 1 m b d m 1l 2 m 2 If we were instead given a distance x, we could caluclate the necessary torque at O and use Newton's Second Law to determine the magnitues of n O and n P . Chapter 12, Number 12 Since the system is in equilibrium the torque about any point must be zero. We can set the pivot to be the edge of the beam that touches the pole. This choice is convenient as it means the torque due to the beam is zero since it acts at the pivot point. Let L represent the length of the beam, which we should be able to eliminate from our final expressions. (a)We apply the rotational analog of Newton's Second Law, being careful to consider the angle between the vectors when taking the cross product. The torque due to the mass at the end is given by
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phy207_hw7_solution - Chapter 12, Number 2 The conditions...

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