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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 F F x F y W R x R R y 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 = 0 3. R x- F x cos = 0 4. F x = 0 5. F x cos - R x sin = 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 + F y sin - W cos = 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 2- F x cos - F y cos = 0 2. 2 F y sin - W cos - 2 F x sin = 0 3. F y cos - W sin + F x sin = 0 4. F y sin - W 2 sin - F x sin = 0 5. F y cos - W 2 cos - F x sin = 0 correct Explanation: For rotational equilibrium about point O, the net torque on the system about that point must vanish. The angle appears as follows x y F x F y W W cos F y cos F x sin 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 F y cos - 2 W cos - F x sin = 0 . Question 4 part 1 of 1 10 points A uniform beam of weight W b and length 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 2 parenrightbigg ( W 1 + W b ) 2. x = ( W 1 + W b ) d + W 2 parenleftbigg 2 parenrightbigg ( W 1 + W 2 ) 3. x = ( W 1 + W b ) d + W 1 parenleftbigg 2 parenrightbigg W 2 correct 4. x = ( W 1 + W b ) d + W 2 parenleftbigg 2 parenrightbigg W 2 5. x = ( W 1 + W b ) d + W 2 parenleftbigg 2 parenrightbigg W 1 6. x = ( W 1 + W 2 ) d + W 1 parenleftbigg 2 parenrightbigg W 2 7. x = ( W 1 + W 2 ) d + W 1 parenleftbigg 2 parenrightbigg ( W 1 + W b ) 8. x = ( W 2 + W b ) d + W 1 parenleftbigg 2 parenrightbigg W 2 Explanation: Basic Concepts: In equilibrium, summationdisplay vector F = 0 summationdisplay vector = 0 Solution: Take the torques about the point P ....
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