{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

qz2sol_3706Hs11

# qz2sol_3706Hs11 - nd law applied to the vertical direction...

This preview shows page 1. Sign up to view the full content.

TA: Tomoyuki Nakayama Tuesday, February 8, 2011 PHY 2048: Physic 1, Discussion Section 3706H Quiz 2 (Homework Set #4) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ A 20.0-kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 50.0 N and a vertical force P are then applied to the block. The coefficient of friction for the block and surface are μ s = 0.400 and μ k = 0.200. a) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 60.0 N. Newton’s 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nd law applied to the vertical direction yields 0 = P + N – mg ⇒ N = mg – P = 136 N The maximum static friction is f s,max = μ s N = 54.4 N The magnitude of horizontal force F is smaller than the maximum static friction. Therefore, the static friction keeps the block from sliding. Newton’s 2 nd law applied to the horizontal direction yields the friction: 0 = F – f f = 50.0 N b) Determine the magnitude of the frictional force acting on the block if the magnitude of P is 80.0 N. The maximum static friction is f s,max = μ s N = μ s (mg – P)= 46.4 N This is smaller than the applied force. Hence the static friction cannot hold the block and the block starts to slide. The friction exerted on the block is kinetic friction. The magnitude of the friction is f = μ k N = μ k (mg – P)= 23.2 N...
View Full 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