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Chapter 04, Problem 07}Onk two {orces &cL Onobject (mass 2.49 kg),te oraxino Find tne (n) mogntude and (6r Jirecoon [relatvethe bccelefauanthe object60-0nJodAmc...

Question

Chapter 04, Problem 07}Onk two {orces &cL Onobject (mass 2.49 kg),te oraxino Find tne (n) mogntude and (6r Jirecoon [relatvethe bccelefauanthe object60-0nJodAmca0o) Numbl

Chapter 04, Problem 07} Onk two {orces &cL On object (mass 2.49 kg), te oraxino Find tne (n) mogntude and (6r Jirecoon [relatve the bccelefauan the object 60-0n JodA mca 0o) Numbl



Answers

Figure $\mathrm{P} 4.21$ shows an object's acceleration-versus-force graph. What is the object's mass?

So in this question it is given to us that there is a certain particle that has a rate of 26th Newton at a place where the acceleration due to gravity G Yes, 9.8 meters. Because they're gonna squared Now we know that the weight equals the mass times acceleration due to gravity. So the mass of this particle would be the weight Do I did my G. And if you plug in the numbers This comes out to be 265 calabria. Yeah. We were asked to calculate the weight and mass of this particle at a place where the acceleration due to gravity. I'm going to write as G one is 4.60 Middles or 2nd squared. We know that the mass uh huh remains constant and does not depend on the value of G. So the mass of this particle At the place where Jeeva is 4.6, 0 would remain the same. 2.65 kg. And the weight would be equals to mass times G one And this would be equal to 12 1919. These are the answers to the first part. Next. It is said that this particle has moved to a place where there is no gravitational force. So G two in this case would be zero. Meadows was thick in the square so the mass once again will remain the same, Which is equals to 2.65 kg. Mhm And the weight in this case would be mass times G2 Which would be equal to zero Newton. These are the answers for the big part.

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.

That the actual weight of the object is equaling 5.0 Newton's and we can say that the apparent weight of the objects F sub G apparent is equaling 3.50 Newton's. We can then say that the mass of the object is equaling 5.0 Newton's divided by G, the acceleration due to gravity. And we can then say that the system we know is at Trent has translational equilibrium in the UAE direction so that some of the forces in the UAE direction will equal zero. This will be the bullion force plus the apparent gravitational force of the apparent weight minus the actual weight. And so we know that again this is all equaling zero so quickly we can see that the buoyant force is equaling 1.50 Newtons. We know that the brilliant forces equaling the density of water multiplied by G times, the volume of the object and so we can then solve for the volume of the object being equal to one rather be consisted of brilliant force divided by the density of water multiplied by G. We can then say that the density of the object would be equaling by definition the mass of the object divided by the volume of the object. And so this would be equaling 5.0 Newton's divided by 9.80 meters per second squared. This would be multiplied by 1000 kilograms per cubic meter. The density of water times 9.80 meters per second squared. This would be divided by 1.50 Newtons and we find that the density of the object is found to be 3.33 times 10 to the third kilograms per cubic meter. This would be our final answer. That is the end of the solution. Thank you for watching.

So for this problem were given a graph which shows a force versus acceleration graph for three objects pulled by rubber bands and were given the mass for object to. And we're trying to find the masses for object one and three. So first, we just need to start with Newton's second law, which is force is equal to m A. And we know. So we need to figure out a so we can use that to solve for the masses. So using the information we know from number two, we have 5.5 for the force on a photograph. Let's go to 5.5. We're just solving for a It's like me. Okay, More than four. And it just works the next objects we have. Yeah, the force on one. We've been estimated about 2.25 It was one of us for this, and I am mhm. The acceleration is six. A size 22 5 equals 16. Solving for em, you're 96. Okay. And then for the last one, you can estimate the force to be 6.25 and the acceleration to be 2.5. Yeah, yeah. Mhm. Yes, It's just plugging in and solving again. We're plugging in this a into these places. So then, for this time as we get about 14, Yeah. So, to this mhm the masses on object, one would be around 96 the mass of the third object would be 14.


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