{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

final ans

# final ans - Graham Andrew – Finalexam 1 – Due noon –...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Graham, Andrew – Finalexam 1 – Due: Dec 15 2005, noon – Inst: Maxim Tsoi 1 This print-out should have 40 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Given: g = 9 . 8 m / s 2 . Consider a boxcar accelerating up a 12 . 4 ◦ 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 angle θ = 19 . 06 ◦ from the perpendicular to the boxcar’s ceiling and floor. a m 1 9 . 6 ◦ 12 . 4 ◦ Given the angles θ = 19 . 06 ◦ > φ = 12 . 4 ◦ , calculate the boxcar’s acceleration a . Correct answer: 1 . 2025 m / s 2 . 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 difference between the true gravity and the inertial forces, so in a non-inertial frame there is a net effective grav- ity ~g eff = ~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 effective gravity (2) must be opposite to the string’s pull on the object, which is 12 . 4 ◦ from the perpendicular to the boxcar’s floor and ceiling and 19 . 06 ◦- 12 . 4 ◦ = 6 . 66 ◦ 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 , find 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 floor while the y axis is perpendicular the the floor: In these coordinates, a x = a, a y = 0 , (3) g x =- g sin φ, g y =- g cos φ (4) where φ = 12 . 4 ◦ is the hill’s slope. At the same time, the string’s direction indicates g eff x =- g eff sin θ, g eff y =- g eff cos θ (5) where θ = 19 . 06 ◦ is the angle between the string and the y axis. Consequently, tan θ = g eff x g eff 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) = 1 . 2025 m / s 2 . 002 (part 1 of 1) 10 points Many satellites orbit the earth about 1036 km above the earth’s surface. Geosynchronous satellites orbit at a distance of 4 . 22 × 10 7 m from the center of the earth....
View Full Document

{[ snackBarMessage ]}

### Page1 / 18

final ans - Graham Andrew – Finalexam 1 – Due noon –...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online