Physics Solution Manual for 1100 and 2101

6 104 n 454 rotational dynamics b the force exerted

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Unformatted text preview: 1.20 ×103 N 23. SSM REASONING The drawing shows the forces acting on the board, which has a length L. The ground exerts the vertical normal force V on the lower end of the board. The maximum force of static friction has a magnitude of µsV and acts horizontally on the lower end of the board. The weight W acts downward at the board's center. The vertical wall applies a force P to the upper end of the board, this force being perpendicular to the wall since the wall is smooth (i.e., there is no friction along the wall). We take upward and to the right as our positive directions. Then, since the horizontal forces balance to zero, we have µsV – P = 0 P V θ W µ sV (1) The vertical forces also balance to zero giving V–W=0 (2) 450 ROTATIONAL DYNAMICS Using an axis through the lower end of the board, we express the fact that the torques balance to zero as L (3) PL sin θ − W cos θ = 0 2 Equations (1), (2), and (3) may then be combined to yield an expression for θ . SOLUTION Rearranging Equation (3) gives tan θ = W 2P (4) But, P = µsV according to Equation (1), and W = V according to Equation (2). Substituting these results into Equation (4) gives tan θ = V 2 µsV = 1 2 µs Therefore, 1 1 = tan −1 2(0.650) = 37.6° 2 µs θ = tan −1 24. REASONING When the wheel is resting on the ground it is in equilibrium, so the sum of the torques about any axis of rotation is zero ( Στ = 0 ) . This equilibrium condition will provide us with a relation between the magnitude of F and the normal force that the ground exerts on the wheel. When F is large enough, the wheel will rise up off the ground, and the normal force will become zero. From our relation, we can determine the magnitude of F when this happens. SOLUTION The free body diagram shows the forces acting on the wheel: its weight W, the normal force FN, the horizontal force F, and the force FE that the edge of the step exerts on the wheel. We select the axis of rotation to be at the edge of the step, so that the torque produced by FE is zero. Letting r FN F FE h W l N , l W , and l F represent the lever arms for the forces FN, W, and F, the sum of the torques is Στ = − FN r2 − (r − h) + W r2 − (r − h) − F (r − h) = 0 13 2 144 244 3 144 244 3 2 lN Solving this equation for F gives Axis of Rotation 2 lW lF Chapter 9 Problems (W − FN ) F= r2 − (r − h) 451 2 r −h When the bicycle wheel just begins to lift off the ground the normal force becomes zero (FN = 0 N). When this happens, the magnitude of F is (W − FN ) F= r 2 − ( r − h) r−h 2 = ( 25.0 N − 0 N ) ( 0.340 m )2 − ( 0.340 m − 0.120 m )2 0.340 m − 0.120 m = 29 N 25. REASONING The drawing shows the beam and the five forces that act on it: the horizontal and vertical components Sx and Sy that the wall exerts on the left end of the beam, the weight Wb of the beam, the force due to the weight Wc of the crate, and the tension T in the cable. The beam is uniform, so its center of gravity is at the center of the beam, which is where its weight can be assumed to act. Since the beam is in equilibrium, the sum of the torques about any axis of rotation must be zero ( Στ = 0 ) , and the sum of the forces in the ( ) horizontal and vertical directions must be zero ΣFx = 0, ΣFy = 0 . These three conditions will allow us to determine the magnitudes of Sx, Sy, and T. T +y 50.0° +τ 30.0° +x Wc Sy Axis Wb 30.0° Sx SOLUTION a. We will begin by taking the axis of rotation to be at the left end of the beam. Then the torques produced by Sx and Sy are zero, since their lever arms are zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, T, in it. Setting the sum of the torques produced by the three forces equal to zero gives (with L equal to the length of the beam) 452 ROTATIONAL DYNAMICS Στ = −Wb ( 1 L cos 30.0°) − Wc ( L cos 30.0° ) + T ( L sin 80.0°) = 0 2 Algebraically eliminating L from this equation and solving for T gives T= = Wb ( 1 cos 30.0°) + Wc ( cos 30.0°) 2 sin 80.0° (1220 N ) ( 1 cos 30.0° ) + (1960 N )( cos 30.0° ) 2 sin 80.0° = 2260 N b. Since the beam is in equilibrium, the sum of the forces in the vertical direction is zero: Σ Fy = + S y − Wb − Wc + T sin 50.0° = 0 Solving for Sy gives S y = Wb + Wc − T sin 50.0° = 1220 N + 1960 N − ( 2260 N ) sin 50.0° = 1450 N The sum of the forces in the horizontal direction must also be zero: Σ Fx = + S x − T cos 50.0° = 0 so that S x = T cos 50.0° = ( 2260 N ) cos 50.0° = 1450 N 26. REASONING If we assume that the system is in equilibrium, we know that the vector sum of all the forces, as well as the vector sum of all the torques, that act on the system must be zero. The figure below shows a free body diagram for the boom. Since the boom is assumed to be uniform, its weight WB is located at its center of gravity, which coincides with its geometrical center. There is a tension T in the cable that acts at an angle θ to the horizontal, as shown. At the hinge pin P, there are two forces acting. The vertical force V that acts o...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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