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The mobile in Fig. 74 is in equilibrium. Object $B$ has mass of0.748 $\mathrm{kg}$ . Determine the masses of objects $\mathrm{A}, \mathrm{C},$ and $\mathrm{D}$ .(Neglect the weights of the crossbars.)

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$0.249 \mathrm{kg}$

Physics 101 Mechanics

Chapter 12

Static Equilibrium; Elasticity and Fracture

Equilibrium and Elasticity

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04:12

In physics, potential energy is the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors. The unit for energy in the International System of Units is the joule (J). One joule can be defined as the work required to produce one newton of force, or one newton times one metre. Potential energy is the energy of an object. It is the energy by virtue of an object's position relative to other objects. Potential energy is associated with restoring forces such as a spring or the force of gravity. The action of stretching the spring or lifting the mass is performed by a force which works against the force field of the potential. The potential energy of an object is the energy it possesses due to its position relative to other objects. It is said to be stored in the field. For example, a book lying on a table has a large amount of potential energy (it is said to be at a high potential energy) relative to the ground, which has a much lower potential energy. The book will gain potential energy if it is lifted off the table and held above the ground. The same book has less potential energy when on the ground than it did while on the table. If the book is dropped from a height, it gains kinetic energy, but loses a larger amount of potential energy, as it is now at a lower potential energy than before it was dropped.

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we have the bottom bar here. We know that each far isn't static equilibrium. So in order to five excuse me To find force sub CD, we can apply the sum of torque. This will equal zero and this will equal m sub d ghee ak sub d minus m sub sea g x subsea and so am subsea with the equaling m sub d times except D over X subsea. This is equaling M sub d multiplied by 17.50 centimeters, divided by 5.0 centimeters. This is giving us 3.50 um 70 and so we can apply the sum of forces in the UAE direction. This system has translational equilibrium in the UAE direction so this is gonna equal zero and we can say that this is equaling force of CD minus m sub sea times G minus m sub d times G. And so we find that force sub CD is equally 4.50 I'm sub d times g and we can keep it. We think you just thio keep in mind this relationship. We then have the middle bar here again. We're gonna ply static equilibrium some of torques is equaling zero. This will equal force subsidy accepts CD minus m sabi G x sub peed. This is rather refined. That forced subsidy is then equaling m sub b g times x sub be divided by X sub CD and we can say that here 4.50 m sub d g equally s a b g X, so be divided by ex subsidy. And so we can say that then M sub deed is equaling. I'm so be X sub be divided by 4.50 times x subsidy and we can solve. So this would be 0.748 kilograms multiplied by 5.0 centimeters, divided by 4.50 multiplied by 15.0 centimeters. And so we find that mass sub d. This is equaling point zero 554 kilograms. We can then say that I'm subsidies equaling 3.50 times. I'm sub d. This is equaling 3.50 times 0.55 for this is equaling 0.194 kilograms. So we find mass subs d here mass subsea here and then we can apply the sum of forces in the UAE direction person. The forces in the war direction equals force of B C D. Minus force of CD times see minus m sub be G Now here we can say that then my apologies. That's not This is just the sun this forces so we don't have to multiply by any length force of CD and then minus m sub DJ. And so this is giving us that this We know that this is equaling zero because we have translational equilibrium in the UAE direction and so force sub B c d. This is equaling essentially 4.50 m sub d plus I'm so be times G and so we can't just keep notes of this relationship. Now we have the top bar here and we can say that here, applying the sum of torques again this equal zero, this would be m sub a times G x sub, a finest force sub B c d times, x sub, B, C d. And so massive massive a wood equaling 4.5 m sub d plus I'm sub b times g x sub B c d. This would be all divided by G X seven a and so and someday is found to be 4.50 I'm sub d plus, um, so be it. The gravity cancels out and this would be multiplied by X sub B C d. Divided by X sub A. And so we can solve. This would be 4.50 multiplied by 0.554 plus 0.748 units. Here, kilograms multiplied by 7.50 centimeters, divided by 30 0.0 centimeters. And we find that massive eh is equaling 0.249 kilograms. This would be our final answer. That is the end of the solution. Thank you for once.

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