Physics 206 Example Problems Newton's Laws of Motion
Physics 206 Example Problems Newton's Laws of Motion
Physics 206 Example Problems Newton's Laws of Motion
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
FH<br />
FG<br />
F<br />
S , K<br />
N<br />
x<br />
y<br />
The various forces have the forms,<br />
N = Nyˆ<br />
FH<br />
= − FH cos( 30 ◦ ) xˆ − FH sin( 30 ◦ ) yˆ<br />
FG<br />
= m gE sin( 30 ◦ ) xˆ − m gE cos( 30 ◦ ) yˆ<br />
F<br />
S , K = FS , K xˆ could point up or down incline<br />
A) F<br />
G = m gE sin( 30 ◦ ) xˆ − m gE cos( 30 ◦ ) yˆ = ( 5) ( 9. 8 N) sin( 30 ◦ ) xˆ − ( 5) ( 9. 8 N) cos( 30 ◦ ) yˆ<br />
F<br />
G = ( 24. 5 N) xˆ − ( 42. 4 N) yˆ<br />
B) Applying the 2nd law in the y direction:<br />
Fnet, y = m a y<br />
− FH sin( 30 ◦ ) + N − m gE cos( 30 ◦ ) = 0<br />
N = FH sin( 30 ◦ ) + m gE cos( 30 ◦ ) = 20 sin( 30 ◦ ) + 42. 4 N = 52. 4 N<br />
C) The static frictional force can supply any force in the range:<br />
− µS N FS µS N<br />
− ( 0. 45) ( 52. 4 N) FS ( 0. 45) ( 52. 4 N)<br />
− 23. 58 N FS 23. 58 N<br />
Let’ s see if the static frictional force can keep the object at rest:<br />
Fnet, x = m ax = 0<br />
m gE sin( 30 ◦ ) + FS − FH cos( 30 ◦ ) = 0<br />
FS = − m gE sin( 30 ◦ ) + FH cos( 30 ◦ ) = − 24. 5 N + 20 cos( 30 ◦ ) N<br />
FS = − 7. 1 8 N<br />
This force is in the range that the static frictional force can supply, so if the object starts at rest it will<br />
remain at rest.<br />
Problem 7. Consider the figure below. Let the coefficient <strong>of</strong> static friction between the block and the<br />
incline be 0. 4.<br />
A) What range <strong>of</strong> masses can the hanging block have if the system is to be in static equilibrium?<br />
B) What is the range <strong>of</strong> tensions that the string will have if the system is in static equilibrium<br />
m1<br />
4<br />
m2 = 10kg<br />
θ = 30 o