Homework 12 - homework 12 BAUTISTA ALDO Due 4:00 am Version...

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homework 12 – BAUTISTA, ALDO – Due: Feb 20 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Given: g = 9 . 8 m / s 2 . Consider a boxcar accelerating up a 15 . 9 slope. Inside the boxcar, an object of un- known mass m hangs on a string attached to the boxcar’s ceiling. When the car accelerates uphill at a steady rate a , the string hangs at a constant an- gle θ = 30 . 4 from the perpendicular to the boxcar’s ceiling and ±oor. a m 30 . 4 15 9 Given the angles θ = 30 . 4 > φ = 15 . 9 , calculate the boxcar’s acceleration a . Correct answer: 2 . 84485 m / s 2 (tolerance ± 1 %). Explanation: Basic Concept: In a non-inertial frame such as an accelerating boxcar, the inertial force - m~a combines with the true gravitational force m~g into a single apparent weight force ~ W app = m ( ~g - ~a frame ) . (1) In fact, the Equivalence Principle says that there is no observable di²erence between the true gravity and the inertial forces, so in a non-inertial frame there is a net efective grav- ity ~g eF = ~g - ~a frame . (2) Solution: Consider the hanging object in the non-inertial frame of the accelerating boxcar. In the car’s frame, the objects hangs without motion so its apparent weight (1) must be balanced by the string’s tension. Hence, the direction of the efective gravity (2) must be opposite to the string’s pull on the object, which is 15 . 9 from the perpendicular to the boxcar’s ±oor and ceiling and 30 . 4 - 15 . 9 = 14 . 5 from the true vertical. At this point, the problem reduces to ge- ometry: Given the directions of vectors ~g , ~a and ~g - ~a and the magnitude g = 9 . 8 m / s 2 , ³nd the magnitude a . We can solve this ques- tion using the sine theorem, but it is just as easy to solve in Carthesian coordinates. Let the x axis run uphull along the boxcars’s ±oor while the y axis is perpendicular the the ±oor: In these coordinates, a x = a, a y = 0 , (3) g x = - g sin φ, g y = - g cos φ (4) where φ = 15 . 9 is the hill’s slope. At the same time, the string’s direction indicates g eF x = - g eF sin θ, g eF y = - g eF cos θ (5) where θ = 30 . 4 is the angle between the string and the y axis. Consequently, tan θ = g eF x g eF y = g x - a x g y - a y = - g sin φ - a - g cos φ = tan φ + a g cos φ . (6) and therefore a = (tan θ - tan φ ) g cos φ (7) = 2 . 84485 m / s 2 . Question 2
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Homework 12 - homework 12 BAUTISTA ALDO Due 4:00 am Version...

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