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A person with weight 900 N stands d = 4.00 m away from the wall on al = 6.00 m beam, as shown in thi beam is 2,000 N. Define upward as the positive y-direction and to the right as the positive x-direction.

A person with weight 900 N stands d = 4.00 m away from the wall on al = 6.00 m beam, as shown in thi beam is 2,000 N. Define upward as the positive y-direction and to the right as the positive x-direction.

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Solve Equation (1) for R (in N) and substitute numerical values: R = N Finalize The positive value for the angle 0 indicates that our estimate of the direction of R was accurate. Had we selected some other axis for the torque equation, the solution might differ in the details ---Select--- . For example, had we chosen an axis through the center of gravity of the beam, the torque equation would involve both T and R. This equation, coupled with Equations (1) and (2), however, could still be solved for the unknowns. Try it! EXERCISE A person with weight 900 N stands d = 4.00 m away from the wall on a { = 6.00 m beam, as shown in this figure. The weight of the beam is 2,000 N. Define upward as the positive y-direction and to the right as the positive x-direction. 30° d Hint (a) Find the tension (in N) in the wire. N (b) Find the horizontal component of the hinge force (in N). (Indicate the direction with the sign of your answer.) N (c) Find the vertical component in (N). (Indicate the direction with the sign of your answer.) N Need Help? Read It
Transcribed Image Text:Solve Equation (1) for R (in N) and substitute numerical values: R = N Finalize The positive value for the angle 0 indicates that our estimate of the direction of R was accurate. Had we selected some other axis for the torque equation, the solution might differ in the details ---Select--- . For example, had we chosen an axis through the center of gravity of the beam, the torque equation would involve both T and R. This equation, coupled with Equations (1) and (2), however, could still be solved for the unknowns. Try it! EXERCISE A person with weight 900 N stands d = 4.00 m away from the wall on a { = 6.00 m beam, as shown in this figure. The weight of the beam is 2,000 N. Define upward as the positive y-direction and to the right as the positive x-direction. 30° d Hint (a) Find the tension (in N) in the wire. N (b) Find the horizontal component of the hinge force (in N). (Indicate the direction with the sign of your answer.) N (c) Find the vertical component in (N). (Indicate the direction with the sign of your answer.) N Need Help? Read It
Standing on a Horizontal Beam A uniform horizontal beam with a length of = 9.10 m and a weight of W, = 225N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 45.0° with the beam (see figure). A person of weight W,- 650 N stands a distance d = 2.25 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam. (a) A uniform beam supported by a cable. A person walks outward on the beam. (b) The force diagram for the beam (c) The force diagram for the beam showing the components of RandT. Rsine Tsin ở Reos e w, Tcos d w, SOLUTION Conceptualize Imagine the person in figure (a) moves outward on the beam. It seems reasonable that the farther he moves outward, the larger the torque he applies about the pivot and the Select the tension in the cable must be to balance this torque. Categorize Because the system is at rest, we categorize the beam as a rigid object -Select- Analyze We identify all the external forces acting on the beam: the 225 N gravitational force, the force T exerted by the cable, the force R exerted by the wall at the pivot, and the 650N force that the person exerts on the beam. These forces are all indicated in the force diagram for the beam shown in figure (b). When we assign directions for forces, it is sometimes helpful to imagine what would happen if a force were suddenly removed. For example, if the wall were to vanish suddenly, the left end of the beam would move Select as it begins to fall. This scenario tells us that the wall is not only holding the beam up but is also pressing outward against it. Therefore, we draw the vector R in the direction as shown in figure (b). Figure (c) shows the horizontal and vertical components of T and R. (Use the following as necessary: T, W W d, , t, and e. Do not substitute numerical values; use variables only.) Applying the first condition of equilibrium, substitute expressions for the forces on the beam into component equations from Fext "0 (1) Ls, - R cos(0) – ( |)-. (2) E, - R sin(o) + T sin(w) – W, - ])-. where we have chosen rightward and upward as our positive directions. Because R, T, and 0 are all unknown, we cannot obtain a solution from these expressions alone. (To solve for the unknowns, the number of simultaneous equations must equal the number of unknowns.) Now let's invoke the condition of rotational equilibrium. A convenient axis to choose for our torque equation is the one that passes through the pin connection. The feature that makes this axis so convenient is that the force R and the horizontal component of T both have a moment arm of zero; hence, these forces produce -Select- torque about this axis. Substitute expressions for the torques on the beam into > S iest = 0: - w,a - w-0 This equation contains only T as an unknown because of our choice of rotation axis. Solve for T (in N) and substitute numerical values: w,d + w.) ( sin(e) Rearrange Equations (1) and (2) and then divide: W, + W, - T sin(@) R sin(8) - tan(0) - R cos(e) Solve for 8 (in degrees) and substitute numerical values: O- tan-"* W, -T sin(e) 6)- 0 - tan T cos(e)
Transcribed Image Text:Standing on a Horizontal Beam A uniform horizontal beam with a length of = 9.10 m and a weight of W, = 225N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of = 45.0° with the beam (see figure). A person of weight W,- 650 N stands a distance d = 2.25 m from the wall. Find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam. (a) A uniform beam supported by a cable. A person walks outward on the beam. (b) The force diagram for the beam (c) The force diagram for the beam showing the components of RandT. Rsine Tsin ở Reos e w, Tcos d w, SOLUTION Conceptualize Imagine the person in figure (a) moves outward on the beam. It seems reasonable that the farther he moves outward, the larger the torque he applies about the pivot and the Select the tension in the cable must be to balance this torque. Categorize Because the system is at rest, we categorize the beam as a rigid object -Select- Analyze We identify all the external forces acting on the beam: the 225 N gravitational force, the force T exerted by the cable, the force R exerted by the wall at the pivot, and the 650N force that the person exerts on the beam. These forces are all indicated in the force diagram for the beam shown in figure (b). When we assign directions for forces, it is sometimes helpful to imagine what would happen if a force were suddenly removed. For example, if the wall were to vanish suddenly, the left end of the beam would move Select as it begins to fall. This scenario tells us that the wall is not only holding the beam up but is also pressing outward against it. Therefore, we draw the vector R in the direction as shown in figure (b). Figure (c) shows the horizontal and vertical components of T and R. (Use the following as necessary: T, W W d, , t, and e. Do not substitute numerical values; use variables only.) Applying the first condition of equilibrium, substitute expressions for the forces on the beam into component equations from Fext "0 (1) Ls, - R cos(0) – ( |)-. (2) E, - R sin(o) + T sin(w) – W, - ])-. where we have chosen rightward and upward as our positive directions. Because R, T, and 0 are all unknown, we cannot obtain a solution from these expressions alone. (To solve for the unknowns, the number of simultaneous equations must equal the number of unknowns.) Now let's invoke the condition of rotational equilibrium. A convenient axis to choose for our torque equation is the one that passes through the pin connection. The feature that makes this axis so convenient is that the force R and the horizontal component of T both have a moment arm of zero; hence, these forces produce -Select- torque about this axis. Substitute expressions for the torques on the beam into > S iest = 0: - w,a - w-0 This equation contains only T as an unknown because of our choice of rotation axis. Solve for T (in N) and substitute numerical values: w,d + w.) ( sin(e) Rearrange Equations (1) and (2) and then divide: W, + W, - T sin(@) R sin(8) - tan(0) - R cos(e) Solve for 8 (in degrees) and substitute numerical values: O- tan-"* W, -T sin(e) 6)- 0 - tan T cos(e)
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