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Unformatted text preview: homework 11 – GADHIA, TEJAS – Due: Apr 11 2007, 4:00 am 1 Question 1 part 1 of 3 10 points A uniform rod pivoted at one end “point O ” is free to swing in a vertical plane in a gravitational field. However, it is held in equilibrium by a force F at its other end. x y ℓ F F x F y W R x R R y θ O Force vectors are drawn to scale. What is the condition for translational equilibrium along the horizontal x direction? 1. R x + F x = 0 correct 2. R x F x sin θ = 0 3. R x F x cos θ = 0 4. F x = 0 5. F x cos θ R x sin θ = 0 Explanation: Using the distances, angles and forces de picted in the figure, the condition summationdisplay F x = 0 for translational equilibrium in the x direction gives R x + F x = 0 . Question 2 part 2 of 3 10 points What is the condition for translational equilibrium along the vertical y direction? 1. R y sin θ + F y sin θ W cos θ = 0 2. R y + F y = 0 3. R y F y + W = 0 4. R y + F y W = 0 correct 5. W R y + F y = 0 Explanation: For the equilibrium in the y direction, summationdisplay F y = 0 gives R y + F y W = 0 . Question 3 part 3 of 3 10 points Taking the origin (point O ) as the pivot point, what is the condition for rotational equilibrium? 1. W ℓ 2 F x ℓ cos θ F y ℓ cos θ = 0 2. 2 F y ℓ sin θ W ℓ cos θ 2 F x ℓ sin θ = 0 3. F y ℓ cos θ W ℓ sin θ + F x ℓ sin θ = 0 4. F y ℓ sin θ W ℓ 2 sin θ F x ℓ sin θ = 0 5. F y ℓ cos θ W ℓ 2 cos θ F x ℓ sin θ = 0 correct Explanation: For rotational equilibrium about point O, the net torque on the system about that point must vanish. The angle θ appears as follows x y ℓ F x F y W W cos θ F y cos θ F x sin θ θ θ θ θ O and we see that the forces R x and R y exert no torque on the point O. From the figure we homework 11 – GADHIA, TEJAS – Due: Apr 11 2007, 4:00 am 2 have ℓ F y cos θ ℓ 2 W cos θ ℓ F x sin θ = 0 . Question 4 part 1 of 1 10 points A uniform beam of weight W b and length ℓ has weights W 1 and W 2 at two positions. W 1 W 2 cm x l d O P The beam is resting at two points. For what value of x will the beam be bal anced at P such that the normal force at O is zero? 1. x = ( W 1 + W b ) d + W 2 parenleftbigg ℓ 2 parenrightbigg ( W 1 + W b ) 2. x = ( W 1 + W b ) d + W 2 parenleftbigg ℓ 2 parenrightbigg ( W 1 + W 2 ) 3. x = ( W 1 + W b ) d + W 1 parenleftbigg ℓ 2 parenrightbigg W 2 correct 4. x = ( W 1 + W b ) d + W 2 parenleftbigg ℓ 2 parenrightbigg W 2 5. x = ( W 1 + W b ) d + W 2 parenleftbigg ℓ 2 parenrightbigg W 1 6. x = ( W 1 + W 2 ) d + W 1 parenleftbigg ℓ 2 parenrightbigg W 2 7. x = ( W 1 + W 2 ) d + W 1 parenleftbigg ℓ 2 parenrightbigg ( W 1 + W b ) 8. x = ( W 2 + W b ) d + W 1 parenleftbigg ℓ 2 parenrightbigg W 2 Explanation: Basic Concepts: In equilibrium, summationdisplay vector F = 0 summationdisplay vector τ = 0 Solution: Take the torques about the point P ....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Force, Work

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