hw4 solution

# hw4 solution - Hw#4 solution Chapter 6 No.5(a In the...

This preview shows pages 1–3. Sign up to view the full content.

Hw #4 solution Chapter 6 No.5 (a) In the horizontal plane, the only force that can keep the coin in circular motion is friction. And because the coin is stationary, so the friction is static friction. (b) 22 2 30.0 50.0 / (50.0 / ) 83.3 / 30.0 r rr s s R cm vc m s m s ac Rc m Fm a f m g μ = = == = = m s m Cancel on both sides 2 2 83.3 / 1 / 0.0850 9.8 / 100 rs s r cm s m ag a g ms cm μμ =⇒ = = = No.9 (a) In vertical direction, the net force on the bob should be zero, because the bob won’t move in the vertical direction. So we have

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

View Full Document
2 cos 0 cos / cos 80.0 9.8 / / cos5.00 787 y FT m g Tm g T m g k g m s θ θθ =− = =⇒ = = × = N N j So we can write the tension in vector way: 787sin 5.00 787cos5.00 (68.6 ) (784 ) Ti j N i =+ = + JG G G G G the vertical component :(784 ) Nj G the horizontal component: (68.6 ) Ni G (b) Just consider the horizontal plane. The bob is in uniform circular motion, so the horizontal component works as the radial force: sin5.00 sin5.00/ rr a a T = m s 2 68.6 /80.0 0.857 / r aN k g m == No.19 (a) At point A, gravity and normal force cause the radial force. And the radial force should be upward. Since the gravity is downward, so the normal
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

hw4 solution - Hw#4 solution Chapter 6 No.5(a In the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online