{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# hw21 - Overton Mays Homework 21 Due 4:00 am Inst Turner...

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

Overton, Mays – Homework 21 – Due: Mar 23 2005, 4:00 am – Inst: Turner 1 This print-out should have 13 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 A triangular wedge of height H = 1 . 404 m, base length L = 2 . 34 m and mass M = 11 . 6 kg is placed on a frictionless table. A small block of mass m = 4 . 71 kg is placed on top of the wedge as shown on the picture below: M m All surfaces are frictionless, so the block slides down the wedge while the wedge slides side- wise on the table. By the time the block slides all the way down to the bottom of the wedge, how far does the wedge slide to the right? Correct answer: 0 . 675745 m. Explanation: Consider the wedge and the block as a two- body system. The external forces acting on this system — the weight of the wedge, the weight of the block and the normal force from the table — are all vertical, hence the net hor- izontal momentum of the system is conserved, P wedge x + P block x = const . Furthermore, we start from rest = center of mass is not moving, and therefore the X coordinate of the center of mass will remain constant while the wedge slides to the right and the block slides down and to the left, X cm = mX block + MX wedge m + M = const . Note that only the X coordinate of the cen- ter of mass is a constant of motion: The Y cm accelerates downward because the P y compo- nent of the net momentum is not conserved. Constant X cm means Δ X cm = 0 and there- fore m Δ X block + M Δ X wedge = 0 . Note that this formula does not depend on where the wedge has its own center of mass; as long as the wedge is rigid, its overall dis- placement Δ X wedge is all we need to know. Finally, consider the geometry of the prob- lem: By the time the block slides all the way down, its displacement relative to the wedge is equal to the wedge length L , or rather - L because the block moves to the left of the wedge. In terms of displacements relative to the inertial frame of the table, this means Δ X block - Δ X wedge = - L. Consequently, 0 = m Δ X block + M Δ X wedge = m ( - L + Δ X wedge ) + M Δ X wedge and therefore Δ X wedge = mL m + M = 0 . 675745 m .

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