{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lec11

# Lec11 - Physics 2A Olga Dudko UCSD Physics Lecture 11 Today...

This preview shows pages 1–7. Sign up to view the full content.

Today: Pulleys. Atwood’s Machine Circular motion. Centripetal force. Physics 2A Olga Dudko UCSD Physics Lecture 11

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

View Full Document
Pulley (Atwood’s Machine) Example Example • Two objects with masses 2.00kg (m 1 ) and 6.00kg (m 2 ) are connected by a light string that passes over a frictionless pulley. Determine the acceleration of each mass and the tension in the string. Solution Solution • First, you must de fi ne a coordinate system. • Let’s choose up as positive y.
Solution Solution Next, draw a force diagram for each mass separately. Forces are already broken up into components. So we should apply Newton’s 2nd Law separately to each object. 6kg mass F tension, string on 6kg mass F gravity, Earth on 6kg mass 2kg mass F tension, string on 2kg mass F gravity, Earth on 2kg mass Atwood’s Machine

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

View Full Document
Solution (cont Solution (cont d) d) • Since the string cannot be stretched, a = a 1 = -a 2 . • Also the tensions will have the same magnitude, T • For the 2kg mass (m 1 ) we have: ± F y = m 1 a 1 ± F y = m 2 a 2 T = m 1 a + m 1 g T - m 1 g = m 1 a T - m 2 g = m 2 (-a) T = m 2 g - m 2 a For the 6kg mass (m 2 ) we have: y equate right- hand sides Atwood’s Machine
Solution (cont Solution (cont d) d) • Set T = T and solve for the resulting acceleration. a = (9.80m/s 2 )(6kg - 2kg)/(6kg + 2kg) m 1 a + m 2 a = m 2 g - m 1 g m 1 a + m 1 g = m 2 g - m 2 a a = (9.80m/s 2 )(4kg)/(8kg) = 4.90m/s 2 Plugging in the values we get: a(m 2 + m 1 ) = g(m 2 - m 1 ) a = g(m 2 - m 1 )/(m 2 + m 1 ) Atwood’s Machine

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.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern