Physics 206 Example Problems Newton's Laws of Motion

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$MATH 219 - Differential Equations Quiz 3 Name Solutions Date ...$

FH FG F S , K N x y The various forces have the forms, N = Nyˆ FH = − FH cos( 30 ◦ ) xˆ − FH sin( 30 ◦ ) yˆ FG = m gE sin( 30 ◦ ) xˆ − m gE cos( 30 ◦ ) yˆ F S , K = FS , K xˆ could point up or down incline A) F 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 F G = ( 24. 5 N) xˆ − ( 42. 4 N) yˆ B) Applying the 2nd law in the y direction: Fnet, y = m a y − FH sin( 30 ◦ ) + N − m gE cos( 30 ◦ ) = 0 N = FH sin( 30 ◦ ) + m gE cos( 30 ◦ ) = 20 sin( 30 ◦ ) + 42. 4 N = 52. 4 N C) The static frictional force can supply any force in the range: − µS N FS µS N − ( 0. 45) ( 52. 4 N) FS ( 0. 45) ( 52. 4 N) − 23. 58 N FS 23. 58 N Let’ s see if the static frictional force can keep the object at rest: Fnet, x = m ax = 0 m gE sin( 30 ◦ ) + FS − FH cos( 30 ◦ ) = 0 FS = − m gE sin( 30 ◦ ) + FH cos( 30 ◦ ) = − 24. 5 N + 20 cos( 30 ◦ ) N FS = − 7. 1 8 N This force is in the range that the static frictional force can supply, so if the object starts at rest it will remain at rest. Problem 7. Consider the figure below. Let the coefficient of static friction between the block and the incline be 0. 4. A) What range of masses can the hanging block have if the system is to be in static equilibrium? B) What is the range of tensions that the string will have if the system is in static equilibrium m1 4 m2 = 10kg θ = 30 o
Solution : There is a range of masses because static friction provides, up to its maximum value, the force it needs to maintain equilibrium. Maximum m1 : The maximum value of m1 is obtained by letting static friction take on its maximum value and act down the incline. For m1 I’ll use a coordinate system such that up is +y. For m2 I’ll let + x be parallel to and down the incline and +y be in the direction of the outward normal to the incline. The free body diagrams for the masses look like: In the chosen coordinate systems: Fron1 m1 FG1 Fron1 = Tyˆ FG1 = − m1 gE yˆ Fron2 = − Txˆ FG2 = m2 gE sinθ xˆ − m2 gE cosθ yˆ N = Nyˆ FS = µS Nxˆ Newton’ s second law applied in the y-direction on mass 1 gives, Fron2 m2 θ FG2 T − m1 gE = m1 a1 y = 0 ⇒ T = m1 gE Newton’ s second law applied in the y-direction on block 2 gives, Newton’ s second law applied in the x-direction on mass 2 gives, Using the results of Equations 1 and 2 in Equation 3 gives, Solving for m1 gives, N FS y x N − m2 gE cosθ = 0 ⇒ N = m2 gE cosθ ( 2) µS N + m2 gE sinθ − T = 0 ( 3) µS m2 gE cosθ + m2 gE sinθ − m1 gE = 0 m1 = µs m2 cosθ + m2 sinθ = { ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 4) m1 , m ax = 8. 46 kg Minimum m1 : The minimum value of m1 is obtained by setting the static frictional force to its maximum value but letting it act up the incline so that F S = − µS Nxˆ . The development leading to Equation 4 still holds except that a negative sign precedes the term involving µS. That is, m1 = − µs m2 cosθ + m2 sinθ = { − ( 0. 4) ( 1 0) cos( 30 ◦ ) + ( 1 0) sin( 30 ◦ ) } kg ( 5) m1 , m ax = 1 . 54 kg The hanging block may have any mass between 1 . 54 kg and 8. 64 kg and the system will be in static equilibrium. B) Equation ( 1 ) gives the range of tensions as 1 5N to 83N. Problem 8. The coefficient of static friction between a box of mass 1 0 kg and a truck bed is 0. 4. What is the maximum possible acceleration of the truck for which the box does not slide on the truck bed? 5 ( 1 )
Page 1 and 2: Problem 1 . Physics 206 Example Pro
Page 3: 7. Plugging in the numbers: Fpy = (
Page 7 and 8: I have assumed directions for the r
Page 9: Applying the second law in the x-di

Physics 206 Example Problems Newton's Laws of Motion

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