Physics Solution Manual for 1100 and 2101

These two relations will allow us to find max

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Unformatted text preview: downward force of gravity mg. Under the influence of these forces, the chandelier is at rest and, therefore, in equilibrium. Consequently, the sum of the x components as well as the sum of the y components of the forces must each be zero. The figure below shows a quasi-free-body diagram for the chandelier and the force components for a suitable system of x, y axes. Note that the diagram only shows one of the forces of tension; the second and third tension forces are not shown in the interest of clarity. The triangle at the right shows the geometry of one of the cords, where l is the length of the cord, and d is the distance from the ceiling. y θ θ T θ d l x mg We can use the forces in the y direction to find the magnitude T of the tension in any one wire. SOLUTION that Remembering that there are three tension forces, we see from the diagram 202 FORCES AND NEWTON'S LAWS OF MOTION 3T sin θ = mg or T= mg mg mg l = = 3 sin θ 3(d / l) 3d Therefore, the magnitude of the tension in any one of the cords is T= (44 kg)(9.80 m/s2 )(2.0 m) = 1.9 × 102 N 3(1.5 m) ____________________________________________________________________________________________ 64. REASONING The diagram at the right shows the force F that the ground exerts on the end of a crutch. This force, as F mentioned in the statement of the problem, acts along the crutch and, therefore, makes an angle θ with respect to the vertical. The horizontal and vertical components of this force are also shown. The horizontal component, F sin θ, is the static frictional force that prevents the crutch from slipping on the F sin θ floor, so fs = F sin θ . The largest value that the static frictional F cos θ θ force can have before the crutch begins to slip is then given by fsMAX = F sin θ MAX . We also know from Section 4.9 (see Equation 4.7) that the maximum static frictional force is related to the magnitude FN of the normal force by fsMAX = µs FN , where µs is the coefficient of static friction. These two relations will allow us to find θ MAX . SOLUTION The magnitude of the maximum static frictional force is given by fsMAX = µs FN . But, as mentioned in the REASONING section, fsMAX is also the horizontal component of the force F, so fsMAX = F sin θ MAX . The vertical component of F is F cos θ MAX and is the magnitude FN of the normal force that the ground exerts on the crutch. Thus, we have fsMAX = µs FN { 123 44 F cosθ MAX F sin θ MAX The force F can be algebraically eliminated from this equation, leaving sin θ MAX = µs cos θ MAX or tanθ MAX = µs The maximum angle that a crutch can have is θ MAX = tan −1 ( µs ) = tan −1 ( 0.90 ) = 42° ______________________________________________________________________________ Chapter 4 Problems 203 65. REASONING The toboggan has a constant velocity, so it has no acceleration and is in equilibrium. Therefore, the forces acting on the toboggan must balance, that is, the net force acting on the toboggan must be zero. There are three forces present, the kinetic frictional force, the normal force from the inclined surface, and the weight mg of the toboggan. Using Newton’s second law with the acceleration equal to zero, we will obtain the kinetic friction coefficient. SOLUTION In drawing the free-body diagram for the toboggan we choose the +x axis to be parallel to the hill surface and downward, the +y direction being perpendicular to the hill surface. We also use fk to symbolize the frictional force. Since the toboggan is in equilibrium, the zero net force components in the x and y directions are +y FN fk 8.00º +x mg 8.00º ΣFx = mg sin 8.00° − µ k FN = 0 ΣFy = FN − mg cos8.00° = 0 In the first of these expressions we have used Equation 4.8 for fk to express the kinetic frictional force. Solving the second equation for the normal force FN and substituting into the first equation gives mg sin 8.00° − µ k mg cos 8.00° = 0 14244 4 3 F or µk = sin 8.00° = tan 8.00° = 0.141 cos 8.00° N 66. REASONING The block is in equilibrium in each case. Since the block moves at a constant velocity in each case, it is not accelerating. A zero acceleration is the hallmark of equilibrium. At equilibrium, the net force is zero (i.e., the forces balance to zero), and we will obtain the magnitude of the pushing force by utilizing this fact as it pertains to the vertical or y ( ) direction. We will use Equation 4.9b ΣFy = 0 for this purpose. 204 FORCES AND NEWTON'S LAWS OF MOTION +y +y It is important to note, however, that the direction of the kinetic frictional force is +x +x not the same in each case. The frictional force always opposes the fk relative motion between the surface of θ θ fk the block and the wall. Therefore, when P P the block slides upward, the frictional W W force points downward. When the block Free-body diagram for Free-body diagram for slides downward, the frictional force upward motion of the downward motion of points upward. These directions are block the block shown in the free-body diagrams (not to scale) for the two cas...
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