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\bullet A uniform $8.0 \mathrm{m}, 1500 \mathrm{kg}$ beam is hinged to a wall andsupported by a thin cable attached 2.0 $\mathrm{m}$ from the free end of the beam, as shown in Figure $10.80 .$ The beam is supported at an angle of $30.0^{\circ}$ above the horizontal. (a) Make a free-body diagram of the beam. (b) Find the tension in the cable. (c) How hard does the beam push inward on the wall?

a) See drawingb) 13219 $\mathrm{N}$c) 17981 $\mathrm{N}$

Physics 101 Mechanics

Chapter 10

Dynamics of Rotational Motion

Newton's Laws of Motion

Rotation of Rigid Bodies

Equilibrium and Elasticity

Cornell University

Rutgers, The State University of New Jersey

Hope College

University of Winnipeg

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so here it might be helpful to draw to diagrams. So we have the X axis, the Y axis and the team we have in the directly in the middle is directly in the center of mass is the weight. In this case it will be directly in the middle. This is we have a theory of Ah, a pivot here, or an access whatever you like to call it. And then we have the white component of the force of the hinge. And then we have the ex component of the force of the hinge. Ah, this is an angle of 30 degrees thiss distance right here, from the pivot to the weight to the weight of the center of mass is 4.0 meters. Um, we have a tension force going. Thanks. This my apologies. And this would be considered two teeth of X and of course, T's apply this from the weight to the tension force. This is 2.0 meters and then again here it's 2.0 meters. We also have that this angle here is 30 degrees, but this single here is 10 degrees so not drawn to scale. However, these air the angle measures and we're going to choose counterclockwise to be positive now. Ah, the second drawing would be of the we can say we can call it. This would be a this would be And this would be simply the force. The tension force. Again, This is R 30 degrees. Thiss distance here would be considered six meters and then we have the tension force here. This angle measure here's 40 degrees. We have the length here l and yet here would be considered the pivot. So this would be simply a section of this beam here, so that would be for party. These air, You're too free diet of free body diagrams and then for part B, they're asking us for the force of the tension. So the magnitude of the tension force rather and this would be the net torque equals zero because not rotating. And this is going to be equal to t times six meters times sign of 40 degrees and then minus the wait times four meters times co sign of 30 degrees. So we can say that the tea is going to be equal to the weight so 1,500 kilograms times 9.8. That's the wait times four meters times co. Sign of 30 degrees divided by six meters times a sign of 40 degrees and we find that the tent, the magnitude of the tension force, will be 1,000 rather 13,000 204 Newtons. So I'd be for part B and then for part See, they're asking us for the magnitude of the force exhibited, rather applied by the hinge. So we'll say Fs for Sigma after the sum of forces in the extraction equal zero This is not moving left or right, and we can say the ex component of the hinge minus T coastline of 10 degrees s O. We know that the exit point of the hinge will be equal to t co sign of 10 degrees and this legal 13,204 co sign of 10 degrees and we find that this is going to be equal to 13,000 three Americans sigma eth In the UAE, direction equals zero and this equals that why component of the hinge, Plus he sign of 10 degrees minus the weight. So we can say that h i V equals of the weight times t sign of 10 degrees and this will be equal to 1,500 times 9.8 minus 13,204 sign of 10 degrees. And this is equaling, huh? 12,407 Newtons. So highlight these. Both, um, we're gonna have to find a magnitude of the hinge. So the magnitude of the force magnitude of the hinge and essentially this would be equal to how hard the beam pushes inward on the wall. But due to new due to Newton's third law of Motion, if the hinges exhibiting a force on the on the bar or the beam the beam will, of course, push that have the same opposite and equal magnitude force on the wall due to Newton's third Law of Motion. And this is simply going to be the square root of the sum of the squares. So 13,000 three squared, plus about 12,407 squared all to the 1/2 power, and this is equaling 17,000 973 Newton ts. So this would be again how hard the beam pushes inward on the wall. This will be your final answer for apart. See, that is the end of the solution. Thank you for watching

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