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\bullet A 9.0 $\mathrm{m}$ uniform beam is hinged to a vertical wall and held horizontally by a 5.0 $\mathrm{m}$ cable attached to the wall 4.0 $\mathrm{m}$ above the hinge, as shown in Figure 10.62 .The metal of this cable has a test strength of 1.00 $\mathrm{kN}$ which means that it will break if the tension in it exceeds that amount. (a) Draw a free-body diagram of the beam. (b) What is the heaviest beam that the cable can support with the given configuration? (c) Find the horizontal and vertical components of the force the hinge exerts on the beam.

a) See Drawingb) $5.33 \times 10^{2} \mathrm{N}$c) $6.00 \times 10^{2} \mathrm{N}$

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

Simon Fraser University

Lectures

02:34

In physics, a rigid body is an object that is not deformed by the stress of external forces. The term "rigid body" is used in the context of classical mechanics, where it refers to a body that has no degrees of freedom and is completely described by its position and the forces applied to it. A rigid body is a special case of a solid body, and is one type of spatial body. The term "rigid body" is also used in the context of continuum mechanics, where it refers to a solid body that is deformed by external forces, but does not change in volume. In continuum mechanics, a rigid body is a continuous body that has no internal degrees of freedom. The term "rigid body" is also used in the context of quantum mechanics, where it refers to a body that cannot be squeezed into a smaller volume without changing its shape.

02:21

In physics, rotational dynamics is the study of the kinematics and kinetics of rotational motion, the motion of rigid bodies, and the about axes of the body. It can be divided into the study of torque and the study of angular velocity.

09:34

A 9.00-m-long uniform beam…

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\bullet A uniform beam 4.0…

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One end of a heavy beam of…

04:33

A uniform, 8.0-m, 1150-kg …

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\bullet A uniform $8.0 \ma…

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In Fig. $12-73,$ a uniform…

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One end of a uniform beam …

so party is asking for a free body diagram of the beam. So let's draw that. Let's first are the y axis. And then right here you can draw the X. So let's just say that this beam is along the x axis. Um, again, it's a 9.0 meter uniform Beam this right here. Ah denotes the pivot or the hinge. And going to the right is the ex component of the hinge. And then going down is the white component of the hinge. Going straight down would be the weight of the beam. And right here we have the tension being applied at an angle. Seita going up would of course be t sign of data. And going to the left would of course, be a T co sign of data. Okay. And at this point, we can say that the distance from the pivot to the tension force right here would be three meters and then we can say the distance from the pivot to center of mass for weight will be for 0.5 meters. So exactly in the middle part, Bean is asking us to do with his party, so that would be the full free body diagram of the beam when the beam is along the X axis. And so for part B, they're asking us for the weight of the beam. We know that we're going to say that counterclockwise is positive and we can say that the tension forces one killer didn't or 1,000 Newtons. And we know that the sum of torque is going to be equal to zero, which would be equal to t sign of Seita times three meters minus the weight times 4.5 meters. And we know that the wait is going to be equal to t sign of fada times three divided by 4.5 and this is equaling 1,000 sign of 53.1 degrees times three divided by 40 rather 4.5. And this is equaling 533 Newton's. So this would be the weight of the beam. And then when we want to find thie forces that the hinge applies so it would be essentially the why and the ex component of the hinge we say some of forces in the extraction equals zero. We know this The bar isn't accelerating in any direction. It's completely at rest. So H h this would be the ex component, as you can see, minus t co sign of Fada. And we know that the ex component of the hinge is going to be equal to 1,000 time co sign of 53.1 degrees and this equals 600 0.4 Newtons. So this would be the ex component of the henge and then for the wide component. Some of forces in the UAE direction equals zero, and this is going to be equal to t sign of theater minus the white component of the hinge minus the weight. So the white component of the hinge will be equal to t sign of Fada minus W. And we can solve 1,000 time's sign of 53.1 degrees, minus 533. And this is giving us 267. Newton's equaling the white component. That is the end of the solution. Thank you for watching

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