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\bullet A uniform beam 4.0 $\mathrm{m}$ long and weighing 2500 $\mathrm{N}$ carries a 3500 $\mathrm{N}$ weight 1.50 $\mathrm{m}$ from the far end, as shown in Figure $10.63 .$ It is supported horizontally by a hinge at the wall and a metal wire at the far end (a) Make a free-body diagram of the beam. (b) How strong does the wire have to be? That is, what is the minimum tension it must be able to support without breaking? (c) What are the horizontal and vertical components of the force that the hinge exerts on the beam?

a) See Drawingb) 6875 $\mathrm{N}$c) $5954 \mathrm{N}$ and $2563 \mathrm{N}$

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

University of Winnipeg

Lectures

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Hello Problem. 47 has a uniform being that's four meters long and weighs 25 100 Newtons and I have a 3500 and wait sitting on the end of it and it's supported by a guide wire as well. So let's take a look at what we got here. We know this is a uniform board, so it's wait, located in the center and it is 25 100 Newtons and that would be two meters from the end because it's a four meter board. Now this wait here that it's supporting is 2.5 meters from the end in its 3500 news. And this is our tension wire here at a 30 degree angle and we also know it's attached here. So we've got a horizontal and vertical force that the hinge must bar exert some vertical or horizontal and vertical force. They were asked to find, um, all three of these forces tension, the horizontal and the vertical force. We know it's an equilibrium, so we have three conditions that must be meant. Arnett torques on this or zero our forces in the ex direction are zero in our net force is in the wind direction. So let's look first at our forces in the extraction, we have one that's totally horizontal. The horizontal force we know that would be equal to the horizontal component of the tension. So it's tension. Sign that. Take that back because I'm 30 now we have a few more vertical forces. We have to downward vertical forces and we have to upward are vertical and our vertical component of the tension that we know that our vertical force of the hinge plus our tensions sign 30 must support the two weights 2500 and waas the 3500 new lights who help us out here at 6000. Right over here on our torques, we have our vertical or horizontal since our pivot points right here are not providing torques because our liver arms or zero, but these two weights to provide a clockwise tour more attention in the wire is providing a counterclockwise tour. So our attention in the wire now it's love arm is the perpendicular distance to this force. So that would be our entire length of four and that would be signed 30 and it would be equal to the 2500 Newton force its level on love too. Wasthe e 3500 didn't force in its lever arm not to it's left arm is 2.5. So here we can actually solve for attention. Well, when we fall for attention, you get 68 75. Since which helps us out here we can then put in 68 75 co sign 30 to get horizontal 59 54 mittens and down here are vertical component, then 25 63. Thank you for

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