Physics 107 HOMEWORK ASSIGNMENT #8
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1 Physics 107 HOMEORK ASSIGMET #8 Cutnell & Johnson, 7 th edition Chapter 9: Problems 16, 22, 24, 66, A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00-kg plate of food and a kg cup of coffee. Obtain the force exerted by the thumb and the force exerted by the four fingers. oth forces act perpendicular to the tray, which is being held parallel to the ground. *22 The drawing shows an outstretched arm (0.61 m in length) that is parallel to the floor. The arm is pulling downward against the ring attached to the pulley system, in order to hold the 98- weight stationary. To pull the arm downward, the latissimus dorsi muscle applies the force M in the drawing, at a point that is m from the shoulder joint and oriented at an angle of 29. The arm has a weight of 47 and a center of gravity (cg) that is located 0.28 m from the shoulder joint. Find the magnitude of M. *24 A woman who weighs 500 is leaning against a smooth vertical wall, as the drawing shows. Find (a) the force F (directed perpendicular to the wall) exerted on her shoulders by the wall and the (b) horizontal and (c) vertical components of the force exerted on her shoes by the ground. *66 The drawing shows a bicycle wheel resting against a small step whose height is h = m. The weight and radius of the wheel are = 25.0 and r = m. A horizontal force is applied to the axle of the wheel. As the magnitude of increases, there comes a time when the wheel just begins to rise up and loses contact with the ground. hat is the magnitude of the force when this happens? *68 A wrecking ball (weight = 4800 ) is supported by a boom, which may be assumed to be uniform and has a weight of As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32, and the angle between the boom and the horizontal is 48. Find (a) the tension in the support cable and (b) the magnitude of the force exerted on the lower end of the boom by the hinge at point P.
2 16. REASOIG AD SOUTIO The net torque about an axis through the contact point between the tray and the thumb is Στ = F( m) (0.250 kg)(9.80 m/s 2 )(0.320 m) (1.00 kg)(9.80 m/s 2 )(0.180 m) (0.200 kg)(9.80 m/s 2 )(0.140 m) = 0 F = 70.6, up Similarly, the net torque about an axis through the point of contact between the tray and the finger is Στ = T ( m) (0.250 kg)(9.80 m/s 2 )(0.280 m) (1.00 kg)(9.80 m/s 2 )(0.140 m) (0.200 kg)(9.80 m/s 2 )(0.100 m) = 0 T = 56.4, down 22. REASOIG The arm, being stationary, is in equilibrium, since it has no translational or angular acceleration. Therefore, the net external force and the net external torque 0.28 m acting on the arm are zero. Using the fact m that the net external torque is zero will allow us to determine the magnitude of the force cg Axis M. The drawing at the right shows three forces: M, the 47- weight of the arm acting 29 at the arm s center of gravity (cg), and the force that acts upward on the right end M of the arm. The 98- force is applied to the arm by the ring. It is the reaction force that (0.070 m) sin 29 arises in response to the arm pulling downward on the ring. Its magnitude is 98, because it supports the 98- weight hanging from the pulley system. Other forces also act on the arm at the shoulder joint, but we can ignore them. The reason is that their lines of action pass directly through the axis at the shoulder joint, which is the axis that we will use to determine torques. Thus, these forces have zero lever arms and contribute no torque. SOUTIO The magnitude of each individual torque is the magnitude of the force times the corresponding lever arm. The forces and their lever arms are as follows: Force ever Arm m M (0.069 m) sin m Each torque is positive if it causes a counterclockwise rotation and negative if it causes a clockwise rotation about the axis. Thus, since the net torque must be zero, we see that 98
3 Solving for M gives M ( )( ) ( )( ) M ( ) m m m sin 29 = 0 ( 98 )( 0.61 m) ( 47 )( 0.28 m) ( m) sin 29 = = REASOIG AD SOUTIO a. The net torque about an axis through the point of contact between the floor and her shoes is Στ = ( )(1.10 m)sin F (cos 30.0 )(1.50 m) = 0 F = 212 b. ewton's second law applied in the horizontal direction gives F h F = 0, so F = 212. h c. ewton's second law in the vertical direction gives F v = 0, so F = v REASOIG hen the wheel is resting on the ground it is in equilibrium, so the sum of the torques about any axis of rotation is zero( Σ τ = 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. hen 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. SOUTIO The free body diagram shows the forces acting on the wheel: its weight, the normal force F, the horizontal force F, and the force F E that the edge of the step exerts on the wheel. e select the axis of rotation to be at the edge of the step, so that the torque produced by F E is zero. etting l, l, and l represent the lever F arms for the forces F,, and F, the sum of the torques is F r F E F h Axis of Rotation 2 2 ( ) ( ) ( ) Σ τ = F r r h + r r h F r h = l l l Solving this equation for F gives F
4 F = ( F ) r 2 ( r h) 2 r h hen the bicycle wheel just begins to lift off the ground the normal force becomes zero (F = 0 ). hen this happens, the magnitude of F is 2 ( F ) r ( r h) ( ) ( ) ( ) m m m F = = = 29 r h m m 68. REASOIG 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 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 on the end of the boom prevents the boom from falling down. The horizontal force H that also acts at the hinge pin prevents the boom from sliding to the left. The weight of the wrecking ball (the "load") acts at the end of the boom. θ = 32 T φ θ V θ = 32 P φ = 48 H y applying the equilibrium conditions to the boom, we can determine the desired forces. SOUTIO The directions upward and to the right will be taken as the positive directions. In the x direction we have F = H T cosθ = 0 (1) while in the y direction we have x F = V T sin θ = 0 (2) y
5 Equations (1) and (2) give us two equations in three unknown. e must, therefore, find a third equation that can be used to determine one of the unknowns. e can get the third equation from the torque equation. In order to write the torque equation, we must first pick an axis of rotation and determine the lever arms for the forces involved. Since both V and H are unknown, we can eliminate them from the torque equation by picking the rotation axis through the point P (then both V and H have zero lever arms). If we let the boom have a length, then the lever arm for is cosφ, while the lever arm for is ( / 2)cosφ. From the figure, we see that the lever arm for T is sin( φ θ ). If we take counterclockwise torques as positive, then the torque equation is cosφ τ = cosφ + T sin ( φ θ ) = 0 2 Solving for T, we have T = cosφ (3) sin( φ θ ) a. From Equation (3) the tension in the support cable is T 1 2 (3600 ) = cos 48 = 1.6 sin(48 32 ) b. The force exerted on the lower end of the hinge at the point P is the vector sum of the forces H and V. According to Equation (1), and, from Equation (2) ( ) 4 4 H = T cosθ = 1.6 cos 32 = 1.4 ( ) 4 4 V = + + T sin θ = sin 32 = 1.7 Since the forces H and V are at right angles to each other, the magnitude of their vector sum can be found from the Pythagorean theorem: FP = H + V = (1.4 ) + (1.7 ) = 2.2
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