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For the hinged beam and cable in example 9-6, what would happen to the tension in the cable as the angle was increased? Would the tension increase, decrease or stay the same? Show your reasoning.

For the hinged beam and cable in example 9-6, what would happen to the tension in the cable as the angle was increased? Would the tension increase, decrease or stay the same? Show your reasoning.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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7. For the hinged beam and cable in example 9-6, what would happen to the tension in the cable as the angle was increased? Would the tension increase, decrease or stay the same? Show your reasoning.
Transcribed Image Text:7. For the hinged beam and cable in example 9-6, what would happen to the tension in the cable as the angle was increased? Would the tension increase, decrease or stay the same? Show your reasoning.
K Hinge FH 7911 L. mg FT Avalonel FTX FIGURE 9-10 Example 9-6. AFTY Mg Arlene's Book Store EXERCISE D Return to the Chapter-Opening Que now. Try to explain why you may have answered differently the first time. I is supported by a cable or cord (Fig. 9-10). It is important to remember that a Our next Example involves a beam that is attached to a wall by a hinge and flexible cable can support a force only along its length. (If there were a compo- flexible.) nent of force perpendicular to the cable, it would bend because it is But for a rigid device, such as the hinge in Fig. 9-10, the force can be in any direction and we can know the direction only after solving the equations. (The hinge is assumed small and smooth, so it can exert no internal torque on the beam.) EXAMPLE 9-6 Hinged beam and cable. A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a small hinge on a wall as shown in Fig. 9-10, The beam is held in a horizontal position by a cable that makes an angle = 30.0°. mine the components of the force F that the (smooth) hinge exerts on the beam, The beam supports a sign of mass M = 28.0 kg suspended from its end. Deter. and the tension Fr in the supporting cable. APPROACH Figure 9-10 is the free-body diagram for the beam, showing all the forces acting on the beam. It also shows the components of FT and a guess for the direction of F. We have three unknowns, FHx, FHу, and Fr (we are given 8), so we will need all three equations, 2F = 0, ΣF, = 0, ΣT = 0. SOLUTION The sum of the forces in the vertical (y) direction is ΣF, = 0 FHy + Fry - mg - Mg = 0. In the horizontal (x) direction, the sum of the forces is ΣFx = 0 FHX FTX = 0. Στ = 0 -(FHY) (2.20 m) + mg(1.10 m) = 0. (ii) For the torque equation, we choose the axis at the point where Fr and Mg act. Then our torque equation will contain only one unknown, FHy, because the lever arms for Fr, Mg, and FHx are zero. We choose torques that tend to rotate the beam counterclockwise as positive. The weight mg of the (uniform) beam acts at its center, so we have We solve for FHy: 1.10 m 2.20 m Equation (i) above gives FHy = mg (0.500) (25.0 kg) (9.80 m/s²) = 123 N. (iii) Next, since the tension Fr in the cable acts along the cable (0 = 30.0°), we see from Fig. 9-10 that tan = Fry/FTx, or Fry = FTx tan0 = FTx (tan 30.0°). Fry=(m + M)g - FHy Equations (iv) and (ii) give = again FTx Fry/tan 30.0° = 396 N/tan 30.0° FHx = FTX = 686 N. (i) HAPTER 9 Static Equilibrium; Elasticity and Fracture 686 N; (53.0 kg) (9.80 m/s²) - 123 N = 396 N. (iv) The components of FH are FHy = 123 N and FHx = 686 N. The tension in the wire is F₁ = VFx + Fry V(686 N)²+(396 N)² = 792 N.¹ = *Our calculation used numbers rounded off to 3 significant figures. If you keep an extra the numbers in your calculator, you get Fry = 396.5 N, FTX = 686.8 N, and Fr= 793 N, all within the expected precision of 3 significant figures (Section 1-4). digit, or leave la zo V V N C *A E 4 n m A t O e F n la S F SC T To ca th ze X- T Fi TH TH Si (rc NC the
Transcribed Image Text:K Hinge FH 7911 L. mg FT Avalonel FTX FIGURE 9-10 Example 9-6. AFTY Mg Arlene's Book Store EXERCISE D Return to the Chapter-Opening Que now. Try to explain why you may have answered differently the first time. I is supported by a cable or cord (Fig. 9-10). It is important to remember that a Our next Example involves a beam that is attached to a wall by a hinge and flexible cable can support a force only along its length. (If there were a compo- flexible.) nent of force perpendicular to the cable, it would bend because it is But for a rigid device, such as the hinge in Fig. 9-10, the force can be in any direction and we can know the direction only after solving the equations. (The hinge is assumed small and smooth, so it can exert no internal torque on the beam.) EXAMPLE 9-6 Hinged beam and cable. A uniform beam, 2.20 m long with mass m = 25.0 kg, is mounted by a small hinge on a wall as shown in Fig. 9-10, The beam is held in a horizontal position by a cable that makes an angle = 30.0°. mine the components of the force F that the (smooth) hinge exerts on the beam, The beam supports a sign of mass M = 28.0 kg suspended from its end. Deter. and the tension Fr in the supporting cable. APPROACH Figure 9-10 is the free-body diagram for the beam, showing all the forces acting on the beam. It also shows the components of FT and a guess for the direction of F. We have three unknowns, FHx, FHу, and Fr (we are given 8), so we will need all three equations, 2F = 0, ΣF, = 0, ΣT = 0. SOLUTION The sum of the forces in the vertical (y) direction is ΣF, = 0 FHy + Fry - mg - Mg = 0. In the horizontal (x) direction, the sum of the forces is ΣFx = 0 FHX FTX = 0. Στ = 0 -(FHY) (2.20 m) + mg(1.10 m) = 0. (ii) For the torque equation, we choose the axis at the point where Fr and Mg act. Then our torque equation will contain only one unknown, FHy, because the lever arms for Fr, Mg, and FHx are zero. We choose torques that tend to rotate the beam counterclockwise as positive. The weight mg of the (uniform) beam acts at its center, so we have We solve for FHy: 1.10 m 2.20 m Equation (i) above gives FHy = mg (0.500) (25.0 kg) (9.80 m/s²) = 123 N. (iii) Next, since the tension Fr in the cable acts along the cable (0 = 30.0°), we see from Fig. 9-10 that tan = Fry/FTx, or Fry = FTx tan0 = FTx (tan 30.0°). Fry=(m + M)g - FHy Equations (iv) and (ii) give = again FTx Fry/tan 30.0° = 396 N/tan 30.0° FHx = FTX = 686 N. (i) HAPTER 9 Static Equilibrium; Elasticity and Fracture 686 N; (53.0 kg) (9.80 m/s²) - 123 N = 396 N. (iv) The components of FH are FHy = 123 N and FHx = 686 N. The tension in the wire is F₁ = VFx + Fry V(686 N)²+(396 N)² = 792 N.¹ = *Our calculation used numbers rounded off to 3 significant figures. If you keep an extra the numbers in your calculator, you get Fry = 396.5 N, FTX = 686.8 N, and Fr= 793 N, all within the expected precision of 3 significant figures (Section 1-4). digit, or leave la zo V V N C *A E 4 n m A t O e F n la S F SC T To ca th ze X- T Fi TH TH Si (rc NC the
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