Physics Solution Manual for 1100 and 2101

When f is large enough the wheel will rise up off the

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Unformatted text preview: t the maximum speed v at which a vehicle can negotiate a curve of radius r is related to the SSF according to v = rg ( SSF ) . No value is given for the radius of the turn. However, by applying this result separately to the sport utility vehicle (SUV) and to the sports car (SC), we will be able to eliminate r algebraically and determine the maximum speed at which the sports car can negotiate the curve without rolling over. SOLUTION Applying v = rg ( SSF ) to each vehicle, we obtain vSUV = rg ( SSF )SUV and vSC = rg ( SSF )SC Dividing these two expressions gives vSC vSUV = rg ( SSF )SC rg ( SSF )SUV or vSC = vSUV (SSF )SC = (18 m/s ) (SSF )SUV 1.4 = 24 m/s 0.80 18. REASONING Since the wheelbarrow is in equilibrium, the net torque acting on it must be zero: Στ = 0 (Equation 9.2). The magnitude of a torque is the magnitude of the force times the lever arm of the force (see Equation 9.1). The lever arm is the perpendicular distance between the line of action of the force and the axis. A torque that tends to produce a counterclockwise rotation about the axis is a positive torque. SOLUTION The lever arms for the forces can be obtained from the distances shown in the text drawing for each design. Equation 9.1 can be used to obtain the magnitude of each torque. We will then write an expression for the zero net torque for each design. These expressions can be solved for the magnitude F of the man’s force in each case: Chapter 9 Problems Left design 447 bg b gb g b gb g 525 + 60.0 b b N g0.400 mgb N g0.600 mg 189 N b F= = Στ = − 525 N 0.400 m − 60.0 N 0.600 m + F 1.300 m = 0 1.300 m Right design bg b gb g 60.0 b b N g0.600 mg 27.7 N F= = Στ = − 60.0 N 0.600 m + F 1.300 m = 0 1.300 m 19. SSM REASONING AND SOLUTION The net torque about an axis through the contact point between the tray and the thumb is 2 2 Στ = F(0.0400 m) − (0.250 kg)(9.80 m/s )(0.320 m) − (1.00 kg)(9.80 m/s )(0.180 m) 2 − (0.200 kg)(9.80 m/s )(0.140 m) = 0 F = 70.6 N, up Similarly, the net torque about an axis through the point of contact between the tray and the finger is 2 2 Στ = T (0.0400 m) − (0.250 kg)(9.80 m/s )(0.280 m) − (1.00 kg)(9.80 m/s )(0.140 m) 2 − (0.200 kg)(9.80 m/s )(0.100 m) = 0 T = 56.4 N, down 20. REASONING The jet is in equilibrium, so the sum of the external forces is zero, and the sum of the external torques is zero. We can use these two conditions to evaluate the forces exerted on the wheels. SOLUTION a. Let Ff be the magnitude of the normal force that the ground exerts on the front wheel. Since the net torque acting on the plane is zero, we have (using an axis through the points of contact between the rear wheels and the ground) Στ = −W lw + Ff lf = 0 where W is the weight of the plane, and lw and lf are the lever arms for the forces W and Ff, respectively. Thus, 448 ROTATIONAL DYNAMICS Στ = −(1.00 × 106 N)(15.0 m − 12.6 m) + Ff (15.0 m) = 0 Solving for Ff gives Ff = 1.60 × 10 5 N . b. Setting the sum of the vertical forces equal to zero yields ΣFy = Ff + 2Fr − W = 0 where the factor of 2 arises because there are two rear wheels. Substituting the data gives ΣFy = 1.60 × 105 N + 2Fr − 1.00 × 106 N = 0 Fr = 4.20 ×105 N 21. REASONING Since the beam is in equilibrium, the sum of the horizontal and vertical forces must be zero: ΣFx = 0 and ΣFy = 0 (Equations 9.4a and b). In addition, the net torque about any axis of rotation must also be zero: Στ = 0 (Equation 9.2). SOLUTION The drawing shows the beam, as well as its weight W, the force P that the pin exerts on the right end of the beam, and the horizontal and vertical forces, H and V, applied to the left end of the beam by the hinge. Assuming that upward and to the right are the positive directions, we obtain the following expressions by setting the sum of the vertical and the sum of the horizontal forces equal to zero: V H P Beam θ θ W Brace Pin (rotational axis) Horizontal forces P cosθ − H = 0 (1) Vertical forces P sin θ + V − W = 0 (2) Using a rotational axis perpendicular to the plane of the paper and passing through the pin, and remembering that counterclockwise torques are positive, we also set the sum of the torques equal to zero. In doing so, we use L to denote the length of the beam and note that the lever arms for W and V are 1 L and L, respectively. The forces P and H create no 2 torques relative to this axis, because their lines of action pass directly through it. Chapter 9 Problems Torques W − c L hVL = 0 1 2 449 ( 3) Since L can be eliminated algebraically, Equation (3) may be solved immediately for V: 1 V = 2W = 1 2 340 = b Ng 170 N Substituting this result into Equation (2) gives 1 P sin θ + 2 W − W = 0 P= W 340 N = = 270 N 2 sin θ 2 sin 39 ° Substituting this result into Equation (1) yields F W Icosθ − H = 0 Gsinθ J HK 2 H= W 340 N = = 210 N 2 tan θ 2 tan 39 ° 22. REASONING AND SOLUTION The net torque about an axis through the elbow joint is Στ = (111 N)(0.300 m) − (22.0 N)(0.150 m) − (0.0250 m)M = 0 or M =...
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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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