Maurice is 180 cm tall. When he stands on a bathroom scale, it reads 75 kg.
Now Maurice lies down on his back, supporting his body with a bathroom scale under his neck (30 cm from the top of his head) and with another scale under his calf (160 cm from the top of his head).
The scale under Maurice's neck reads 40 kg. What is the location of Maurice's center of mass, measured from the top of his head? (Round to the nearest centimeter, if necessary.)
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Surely, he must be a fit guy, if he can hold his body up in the air like this...I certainly wouldn't be able to this before the two scale even showed a clear reading, I could calculate with.
Also, most body scales have a measurement scale that has at least one more significant number.
Couldn't the result of this problem be exatly 91cm or at least say aproximatly 91cm? I was confused when I got a decimal answer.
Most people have solved using the direct formula; I would like to mention on a more basic (yet longer) method: Let the distance of head from COM(Centre of mass) be X and that of legs be Y X+Y=180-(30+(180-160)) ....... X+Y=130 -1 The Sum of moments of force on COM is 0(by definition of COM and as body is at equilibrium).But there only two forces other than weight(which acts on COM and produces 0 moment due to 0 distance)Therefore X 40=Y (75-40)..........40X-35Y=0 -2 Solving we get 75X=130*35 ; X=60.67 distance of COM from head is X+30 =60.67+30=90.67=91
I reject this and any other solution leading to one of the answers printed as options. Here are two reasons: 1) The scale under his head reads 40kg so the one under his feet must read 35kg (to total 75kg). This fact alone indicates that his center of gravity must be closer to his head than his feet. Since he is 180cm tall, that center of gravity must be something LESS THAN 90cm from his head. All the options presented were 90 or greater, so all incorrect. 2) The fact that he is not spinning around his center of gravity means that 40X = 35(180-X) where X is the distance from the top of head to center of gravity, as the problem specified. 40X+35X=6300, so 75X=6500, so X= 86.67
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Sorry, I made a typo. Should have been 6300/75=84
Proved differently, he has two points of support around which he could rotate. The fact that he is not spinning around the pivot point at top of his head means that the clockwise rotational force (=75X) due to gravity acting through the center of his mass must be exactly offset by the counter clockwise rotational force (=180x35) pushing up on his feet. So, 75X=6300 and thus X=84
Found my error. I thought it had said that scale was at top of head.
Did the same way!!!
exactly same process was done by me
Ans. 35/75*130+30=90.6667 or 91
Most people have solved using the direct formula; I would like to mention on a more basic (yet longer) method: Let the distance of head from COM(Centre of mass) be X and that of legs be Y X+Y=180-(30+(180-160)) ....... X+Y=130 -1 The Sum of moments of force on COM is 0(by definition of COM and as body is at equilibrium).But there only two forces other than weight(which acts on COM and produces 0 moment due to 0 distance)Therefore X 40=Y (75-40)..........40X-35Y=0 -2 Solving we get 75X=130*35 ; X=60.67 distance of COM from head is X+30 =60.67+30=90.67=91
Let m 1 = 3 0 k g , m = 7 5 k g , d 1 = 0 . 3 m , d 2 = 1 . 6 m . We ought to find x , the distance from Maurice's head to his center of mass. There are exactly 3 forces involved in this process: The reaction force, N 1 , which has its origin at d 1 from Maurice's head, N 2 , the other reaction force and F W , the weight of the man. Translational equilibrium tells us that: N 1 + N 2 + F W = 0 ⟹ N 2 = F W − N 1 Applying rotational equilibrium relative to the origin of the graph, Maurice's head: τ N 1 + τ N 2 + τ F W = 0 And therefore: N 1 d 1 + N 2 d 2 = F W x ⟹ m 1 g d 1 + ( m − m 1 ) g d 2 = m g x x = m m 1 d 1 + ( m − m 1 ) d 2
Numerically speaking, this is equivalent to x = 7 5 6 8 m = 9 0 , ( 6 ) c m ≈ 9 1 c m .
Most people have solved using the direct formula; I would like to mention on a more basic (yet longer) method: Let the distance of head from COM(Centre of mass) be X and that of legs be Y X+Y=180-(30+(180-160)) ....... X+Y=130 -1 The Sum of moments of force on COM is 0(by definition of COM and as body is at equilibrium).But there only two forces other than weight(which acts on COM and produces 0 moment due to 0 distance)Therefore X 40=Y (75-40)..........40X-35Y=0 -2 Solving we get 75X=130*35 ; X=60.67 distance of COM from head is X+30 =60.67+30=90.67=91
Using the definition that an objects center of mass is the location where all the mass is averaged out we can take the weighted average of the scale's results. It reads 40 kilograms 30 centimeters from the top of his head so it must weigh 35 kilograms 160 centimeters from the top of his head. 7 5 4 0 represents the fraction of his mass 30 centimeters from his head 7 5 3 5 represents the fraction of his mass 160 centimeters from his head We can simply find the sum of 7 5 4 0 times 30 and 7 5 3 5 times 160 to find the weighted average of his mass. This results in a value of 90 3 2 centimeters which is approximately 91 centimeters.
The sum of the moments about the origin must be = 0. all we need is a single CCW moment arm with 75 kg to balance. Sum of CW moments are 40 x 30 + 35 x 160 = 6800 kg-cm.... 6800 / 75 = 90.666.. or 91 cm
You sir must be a pilot. This is how I solved it also. IMO the simplest and most elegant solution.
Treating Maurice as a homogeneous bar, this is torquewise equivalent to balancing a bar of 130 cm on a fulcrum, with 40 kg and 35 kg on the ends. The torques on the ends must be equal, with the center of mas at a distance d from where the head is supported:
4 0 ⋅ d = 3 5 ( 1 3 0 − d )
This solves to d = 6 0 . 6 7 , so rounding and adding the 30 cm from the top of the head yields 91 cm.
Obs.: There is an implicit a hypothesis that you must treat Maurice as a homogeneous bar.
As a simpleton, I mean pilot I'll show you we use weight & balance daily to solve this exact problem.
WxA=M. (Weight, Arm is the fixed distance from a reference point, Moment is are working variable). Add you moments together and divide by your total weight to get the Arm again, this becomes your center of gravity or center of mass in this case.
40kg*30cm=1200
35kg*160cm=5600
6800÷75=
90.666666
40/75 = 53.33 53.33 * 160 = approx 91. I feel like I lucked out on this one.
Hideous! The answer is 90+2/3. Why should we be rounding without warning before selecting from the multiple choice? Is the poser unable to find numbers thst yield an integer answer? Or does he/she delight in trying to convince the responder that their carefully thought-out logic must be wrong (when it isn’t)?
In science it is common to round answers to a reasonable number of significant figures. Since the relevant data is given with two significant figures, the answer may also be given in that format.
More specifically, the measurements are given in whole centimeters and in whole kilograms, with an uncertainty of approx. 1%. The center of mass can therefore only be determined with an uncertainty of approx. 1%; therefore the answer is rounded to whole integers. It would therefore be irresponsible to specify the answer as 90.7 cm; and certainly to suggest infinite precision by writing 9 0 3 2 cm.
We can solve this problem as if it was a static structure.
Because it is static, we can say that the sum of the moments around any point equals 0.
When we take the sum of the moment around the right scale, we get:
40.g.1,3 = 75.g.X
X = 7 5 4 0 . 1 , 3 = 0,69 cm to the left of the right scale, wich is +/- 91 cm from the top of his head.
Using the definition that an objects center of mass is the location where all the mass is averaged out we can take the weighted average of the scale's results. It reads 40 kilograms 30 centimeters from the top of his head so it must weigh 35 kilograms 160 centimeters from the top of his head. represents the fraction of his mass 30 centimeters from his head represents the fraction of his mass 160 centimeters from his head We can simply find the sum of times 30 and times 160 to find the weighted average of his mass. This results in a value of 90 centimeters which is approximately 91 centimeters.
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The torques on Maurice's body are in equilibrium. There is one clockwise torque from the force of gravity at his center of mass; and two counterclockwise torques from the supporting scales.
For the torque analysis, we can take any point as the reference point; we might as well take the origin x = 0 (top of head) as reference point. Thus we find m g x C M = m 1 g x 1 + m 2 g x 2 , where m 1 , m 2 are the scale readings and x 1 , x 2 the positions of the scales. Cancel g . Note also that m = m 1 + m 2 : together, the two scales support Maurice's weight. x C M = m m 1 x 1 + m 2 x 2 = 7 5 4 0 ⋅ 3 0 + 3 5 ⋅ 1 6 0 ≈ 9 1 c m ; or also x C M = x 2 − m m 1 ( x 2 − x 1 ) = 1 6 0 − 7 5 4 0 ⋅ 1 3 0 ≈ 9 1 c m .