Physics 206 Example Problems Newton's Laws of Motion

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

Solution : If the box is not to slip it must have the same acceleration as the truck. The forces acting on the box are a static frictional force, the normal force and the gravitational pull of the earth on the box. The static frictional force points in a direction so as to maintain no relative motion of the box and the truck. In this case that means that the static frictional force points in the direction of the acceleration since in fact, it is the force responsible for the acceleration of the box. A free body diagram is shown below. In this coordinate system the forces have the form, Applying Newton’ s second law in the y-direction gives, N y x m N = Nyˆ F G = − m gE yˆ F S = µS Nxˆ FG FS N − m gE = 0 ⇒ N = m gE Applying the second law in the x-direction and using the above result gives, Solving for the x-component of the acceleration gives, µS N = m ax ⇒ µS m gE = m ax ax = µS gE = 0. 4 ( 9. 8) m/s 2 = 3. 92 m/s 2 Problem 9. Two blocks are connected by a massless, rigid rod and placed on an inclined plane as shown in the figure below. The blocks have masses m1 and m2 and coefficients of kinetic friction ( with the plane) of µK1 and µK2. A) Find a symbolic formula for the acceleration of the system. B) Find a symbolic expression for the force that the connecting rod exerts on either block. C) Show that the force from part B) is zero when µK1 = µK2. Solution : The free body diagrams for the two blocks looks like, y FK1 m1 θ FG1 N1 Fron1 x m1 Note that I have chosen the usual coordinate systems for inclined planes. In this coordinate system the forces that act on block 1 can be written as, ¡ N ¡ 1 = N1 yˆ F m2 ¡ G1 = m1 gE sinθ xˆ − m1 gE cosθ yˆ F The forces that act on mass 2 can be written as, ¡ N ¡ 2 = N2 yˆ F ¡ G2 = m2 gE sinθ xˆ − m2 gE cosθ yˆ F 6 θ FK2 Fron2 m2 θ ¡ ron1 = Txˆ F FG2 ¡ ron1 = − Txˆ F N2 y x K 1 = − µK1 N1 xˆ K2 = − µK2 N2 xˆ
I have assumed directions for the rod force on the blocks. All that I know for sure is that the forces that the rod exerts on the masses are equal and opposite, so T may come out to be positive or negative. Applying the second law in the y-direction on mass 1 gives, Fnet, y = m1 a1 y = 0 ⇒ N1 − m1 gE cosθ = 0 ⇒ N1 = m1 gE cosθ ( 6) Similarly the second law applied in the y-direction for mass 2 gives, Fnet, y = m2 a2 y = 0 ⇒ N2 − m2 gE cosθ = 0 ⇒ N2 = m2 gE cosθ ( 7) The second law applied in the x-direction for mass 1 gives, T − µK1 N1 + m1 gE sinθ = m1 a1 x The second law applied in the x-direction for mass 2 gives, − T − µK2 N2 + m2 gE sinθ = m2 a1 x A) Note that the two blocks have the same acceleration. Using Equations 6 and 7 in Equations 8 and 9 and then adding 8 and 9 gives, − ( µK1 m1 + µK2 m2 ) gE cosθ + ( m1 + m2) gE sinθ = ( m1 + m2) a1 x Solving for the acceleration component gives, a1 x = − ( µK1 m1 + µK2 m2 ) gE cosθ + ( m1 + m2) gE sinθ m1 + m2 B) Dividing Equation 8 by m1 and Equation 9 by m2 and then taking the difference of the results gives, Simplifying gives, 1 T C) Note that if µK1 = µK2, then T = 0. m1 + 1 m2 − ( µK1 − µK2) gE cosθ = 0 T = ( µK1 − µK2) gE cosθ m1 m2 m1 + m2 Problem 1 0. Your car is stuck in a mud hole. You are alone but have a long, strong rope. Having studied physics you tie the rope tautly to a telephone pole and pull on it sideways as in the figure below. A) Find the forces exerted by the rope on the car when θ = 3 ◦ and you are pulling with a force of 400 N and the car does not budge. B) How strong must the rope be if it takes a force F of 600 N to move the car when θ = 4 ◦ . Car θ θ Solution : A) Apply the second law to the junction at which force the force F is applied. Call F1 the force of the rope attached to the car on this junction and F2 the force of the rope attached to the pole acting on this junction. In the diagram take up to +y and to the right +x. In this coordinate system I can write, F F Pole = − Fyˆ F 1 = − T1 cosθ xˆ + + T1 sinθ yˆ F 2 = T2 cosθ xˆ + T2 sinθ yˆ Applying the second law in the x-direction gives, Fnet, x = m ax − T1 cosθ + T2 cosθ = 0 T1 = T2 ≡ T 7 ( 8) ( 9) ( 1 0) ( 1 1 )
Page 1 and 2: Problem 1 . Physics 206 Example Pro
Page 3 and 4: 7. Plugging in the numbers: Fpy = (
Page 5: Solution : There is a range of mass
Page 9: Applying the second law in the x-di

Physics 206 Example Problems Newton's Laws of Motion

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