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$\bullet$ The farmyard gate. A gate 4.00 $\mathrm{m}$ wide and 2.00 $\mathrm{m}$ high weighs 500 $\mathrm{N} .$ Its center of gravity is at its center, and it is hinged at $A$ and $B .$ To relieve the strain on the top hinge, a wire $C D$ is connected as shown in Fig. $10.72 .$ The tension in $C D$ is increased until the horizontal force at hinge $A$ is zero. (a) What is the tension in the wire $C D ?$ (b) What is the magnitude of the horizontal component of the force at hinge $B ?($ c) What is the combined vertical force exerted by hinges $A$ and $B ?$

a) 268 $\mathrm{N}$b) 232 $\mathrm{N}$c) 366 $\mathrm{N}$

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

University of Winnipeg

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Let's draw the free body diagram. We're going to say that counterclockwise is positive. And so we can say that we're just going to use one whole workbook for the three body diagram. Um, we have here. This is labeled as bi. We're gonna label. This is a the ex component of the pivot at a does not exist. So we simply only have the tinge of a in the UAE the Y component of the force of a and then here age sub B B h syrupy for their age. So b h and we can say that it's strong here we have a tension force and here, of course, is T sub. Why here, of course, is teeth of X. And this angle right here is 30 degrees. Okay. And at this point, we have this distance here, both equalling to 0.0 meters directly in the center of This is the weight. Um, and at this point, we can say that this height over here is also to 0.0.0 meters. So this would be the full free body diagram of of the context here. So for part A, there are asking us for the tension force so we can say that for party, the sum of the torque at point B will be equal to zero will be equal to t sign of 30 degrees times four meters plus TT co sign of 30 degrees times two meters minus the wait Times two t will now will be equal to two times sign of 30 degrees plus co sign of 30 degrees and as a numerator the weight So this will be equal to to sign of 30 degrees plus co sign of 30 degrees on 500 as the numerator and this equals 268 noons. So this will be your answer for a party for part B. They're asking us for the essentially the ex component of the force of the hinge at point B. So we can say that the sum of forces in the ex direction will equal zero. This will equal H b h minus t co sign of 30 degrees s. So we can say that H B H is going to be equal to t co sign of 30 degrees and so we can say age so b h equals 268 Times CO sign of 30 and this is going to give us 232 Nunes. So that's your answer for Part B and then for a part f apart. See, you want the white component of the force of the hinge at Point A and Point B So Sigma F why equals again? Zero. It's not moving over down. This equals ages of a V plus H p V plus T sign of 30 degrees. Sign of 30 is simply one over two, minus the weight. So agents of a V Plus agents be the equals the weight minus T sign of 30 degrees. So this will be equal to 500 minus 2 68 divided by two. And this is giving us 366 noon. So this appearance frontal answer for Part C and again realized that we're finding the sum. Um, it's actually given the scenario in the free body diagram here, it's actually impossible to find these two forces independent of each other. Rather to find these two forces separately. We can only find the sum based on the sum of forces in the white direction. We don't have enough information in order to solve for these two forces independently, that is the end of the solution. Thank you for watching

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