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S E C T I O N 5 . 7 • Some Applications of Newton’s Law 125 Conceptual Example 5.5 Forces Between Cars in a Train Example 5.6 The Runaway Car A car of mass m is on an icy driveway inclined at an angle !, (A) Find the acceleration of the car, assuming that the driveway is frictionless. Solution Conceptualize the situation using Figure 5.11a. From the plane, and the gravitational force F g " m g, which acts vertically downward. For problems involving inclined planes, Now we apply Newton’s second law in component form, noting that a y " 0: (2)
# F y " n # mg
cos ! " 0 (1)
# F x " mg
sin !
" ma x problem, we construct two free-body diagrams—one for the With reference to Figure 5.10b, we apply the equilib- rium condition in the y direction, #F y " 0 : T 3 # F g " 0. This leads to T 3 " F g " 122 N. Thus, the upward force T 3 exerted by the vertical cable on the light balances the gravi- Next, we choose the coordinate axes shown in Figure 5.10c and resolve the forces acting on the knot into their the weight of the light. We solve (1) for T 2 in terms of T 1 to obtain This value for T 2 is substituted into (2) to yield Both of these values are less than 100 N ( just barely for T 2 ), so the cables will not break. Let us finalize this problem by imag- What If? What If? Suppose the two angles in Figure 5.10a are equal. What would be the relationship between T 1 and T 2 ? Answer We can argue from the symmetry of the problem 1 and T 2 would be equal to each other. Mathematically, if the equal angles are called !, Equa- which also tells us that the tensions are equal. Without 1 and T 2 . However, the tensions will be equal to each other, regardless of the value of !. T 2 " T 1 $ cos ! % " T 1 T 2 " 1.33T 1 " 97.4
N T 1 " 73.4
N T 1
sin
37.0' $ (1.33T 1 )(sin
53.0') # 122
N " 0 (3)
T 2 " T 1 $ cos
37.0 ' cos
53.0 ' % " 1.33T 1 Train cars are connected by couplers, which are under tension Solution As the train speeds up, tension decreases from the first car must apply enough force to accelerate the rest When the brakes are applied, the force again decreases from front to back. The coupler connecting the locomotive Force x Component y Component T 1 # T 1 cos 37.0° T 1 sin 37.0° T 2 T 2 cos 53.0° T 2 sin 53.0° T 3 0 # 122 N Knowing that the knot is in equilibrium (a " 0) allows us to (1) (2) From (1) we see that the horizontal components of T 1 and T 2 must be equal in magnitude, and from (2) we see that the sum of the vertical components of T 1 and T 2 must bal- ance the downward force T 3 , which is equal in magnitude to # F y " T 1
sin 37.0' $ T 2
sin 53.0' $ (# 122
N) " 0 # F x " # T 1
cos 37.0' $ T 2
cos 53.0' " 0 |