Mixing
Is it possible to find the mass of an unknown object if the length of a meterstick isn't given?
$begingroup$
I've been going on circles with this problem these days, and although the problem doesn't seem to be that complicated I'm still lost at where is that missing statement or clue to solve this riddle.
The problem is as follows:
In Figure shown below, the meterstick's mass is $textrm{0.160 kg}$
and the string tension is $textrm{2.50 N}$. The system is in
equilibrium. Find (a)the unknown mass $m$ and (b) the upward force the
fulcrum exerts on the stick.
Now what I did in my attempt to solve this problem was to use this fact from Rotational equilibrium:
$$tau_{net}=0$$
Therefore to do so I had to establish the lengths of the weights attached to the meterstick, provided that there are four forces acting.
To which I used the following notation $m_{s}$ for the mass of the meterstick, $m$ mass of the unknown object, $g$ for gravity and $x$ for the unknown length to the center of mass.
From static equilibrium in $textrm{y-axis}$:
$$F_net=0$$
$$F_{1}=textrm{normal force}=mg+m_{s}g+2.5$$
With this established all was to apply the definition given earlier to rotational equilibrium first at the fulcrum:
$tau_{1}= +mgleft(0.3right)$
$tau_{2}= -m_sgleft(xright)$
$tau_{3}= -2.5left(2x+0.30 right)$
Summing all these three and equating to zero I obtained:
$0 = +mgleft(0.3right)-m_sgleft(xright)-2.5left(2x+0.30 right)$
$0.3mg-m_{s}gx-5x-0.75=0$
$$x=frac{0.3mg-0.75}{m_{s}g+5}$$
Then I thought to use the above equation combined with the one obtained from static equilibrium in the y-axis and a new analysis of the rotational equilibrium but this time in the center of mass to relate those and obtain the length in terms of the mass with the hopes of solving the system of equation.
This is shown in the diagram below:
As mentioned earlier the distance to the center of mass:
$tau_{1}= + mg left( 0.3 + xright)$
$tau_{2}= - left(mg+m_{s}g+2.5right)x$
$tau_{3}= - 2.5 left( x+0.3 right)$
By summing these up and equating to zero this becomes into:
$+ mg left( 0.3 + xright) - left(mg+m_{s}g+2.5right)x - 2.5 left( x+0.3 right) = 0$
$0.3 mg + mgx - mgx - m_{s}gx - 2.5x -2.5x -0.75=0$
$-left(5+m_{s}gright)x+0.3 mg -0.75 = 0$
$$frac{0.3 mg -0.75}{5+m_{s}g}=x$$
Which in the end reproduces the same result obtained earlier. Although I don't end up with a contradiction. How am I supposed to obtain the length?. Could it be that did I overlooked something, was it my system of equations wrong? or just something in the analysis that is incomplete?. I hope somebody could guide me with this problem as I'm still stuck on it.
algebra-precalculus systems-of-equations physics
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
|
show 5 more comments
$begingroup$
I've been going on circles with this problem these days, and although the problem doesn't seem to be that complicated I'm still lost at where is that missing statement or clue to solve this riddle.
The problem is as follows:
In Figure shown below, the meterstick's mass is $textrm{0.160 kg}$
and the string tension is $textrm{2.50 N}$. The system is in
equilibrium. Find (a)the unknown mass $m$ and (b) the upward force the
fulcrum exerts on the stick.
Now what I did in my attempt to solve this problem was to use this fact from Rotational equilibrium:
$$tau_{net}=0$$
Therefore to do so I had to establish the lengths of the weights attached to the meterstick, provided that there are four forces acting.
To which I used the following notation $m_{s}$ for the mass of the meterstick, $m$ mass of the unknown object, $g$ for gravity and $x$ for the unknown length to the center of mass.
From static equilibrium in $textrm{y-axis}$:
$$F_net=0$$
$$F_{1}=textrm{normal force}=mg+m_{s}g+2.5$$
With this established all was to apply the definition given earlier to rotational equilibrium first at the fulcrum:
$tau_{1}= +mgleft(0.3right)$
$tau_{2}= -m_sgleft(xright)$
$tau_{3}= -2.5left(2x+0.30 right)$
Summing all these three and equating to zero I obtained:
$0 = +mgleft(0.3right)-m_sgleft(xright)-2.5left(2x+0.30 right)$
$0.3mg-m_{s}gx-5x-0.75=0$
$$x=frac{0.3mg-0.75}{m_{s}g+5}$$
Then I thought to use the above equation combined with the one obtained from static equilibrium in the y-axis and a new analysis of the rotational equilibrium but this time in the center of mass to relate those and obtain the length in terms of the mass with the hopes of solving the system of equation.
This is shown in the diagram below:
As mentioned earlier the distance to the center of mass:
$tau_{1}= + mg left( 0.3 + xright)$
$tau_{2}= - left(mg+m_{s}g+2.5right)x$
$tau_{3}= - 2.5 left( x+0.3 right)$
By summing these up and equating to zero this becomes into:
$+ mg left( 0.3 + xright) - left(mg+m_{s}g+2.5right)x - 2.5 left( x+0.3 right) = 0$
$0.3 mg + mgx - mgx - m_{s}gx - 2.5x -2.5x -0.75=0$
$-left(5+m_{s}gright)x+0.3 mg -0.75 = 0$
$$frac{0.3 mg -0.75}{5+m_{s}g}=x$$
Which in the end reproduces the same result obtained earlier. Although I don't end up with a contradiction. How am I supposed to obtain the length?. Could it be that did I overlooked something, was it my system of equations wrong? or just something in the analysis that is incomplete?. I hope somebody could guide me with this problem as I'm still stuck on it.
algebra-precalculus systems-of-equations physics
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
3
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
2
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
1
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26
|
show 5 more comments
$begingroup$
I've been going on circles with this problem these days, and although the problem doesn't seem to be that complicated I'm still lost at where is that missing statement or clue to solve this riddle.
The problem is as follows:
In Figure shown below, the meterstick's mass is $textrm{0.160 kg}$
and the string tension is $textrm{2.50 N}$. The system is in
equilibrium. Find (a)the unknown mass $m$ and (b) the upward force the
fulcrum exerts on the stick.
Now what I did in my attempt to solve this problem was to use this fact from Rotational equilibrium:
$$tau_{net}=0$$
Therefore to do so I had to establish the lengths of the weights attached to the meterstick, provided that there are four forces acting.
To which I used the following notation $m_{s}$ for the mass of the meterstick, $m$ mass of the unknown object, $g$ for gravity and $x$ for the unknown length to the center of mass.
From static equilibrium in $textrm{y-axis}$:
$$F_net=0$$
$$F_{1}=textrm{normal force}=mg+m_{s}g+2.5$$
With this established all was to apply the definition given earlier to rotational equilibrium first at the fulcrum:
$tau_{1}= +mgleft(0.3right)$
$tau_{2}= -m_sgleft(xright)$
$tau_{3}= -2.5left(2x+0.30 right)$
Summing all these three and equating to zero I obtained:
$0 = +mgleft(0.3right)-m_sgleft(xright)-2.5left(2x+0.30 right)$
$0.3mg-m_{s}gx-5x-0.75=0$
$$x=frac{0.3mg-0.75}{m_{s}g+5}$$
Then I thought to use the above equation combined with the one obtained from static equilibrium in the y-axis and a new analysis of the rotational equilibrium but this time in the center of mass to relate those and obtain the length in terms of the mass with the hopes of solving the system of equation.
This is shown in the diagram below:
As mentioned earlier the distance to the center of mass:
$tau_{1}= + mg left( 0.3 + xright)$
$tau_{2}= - left(mg+m_{s}g+2.5right)x$
$tau_{3}= - 2.5 left( x+0.3 right)$
By summing these up and equating to zero this becomes into:
$+ mg left( 0.3 + xright) - left(mg+m_{s}g+2.5right)x - 2.5 left( x+0.3 right) = 0$
$0.3 mg + mgx - mgx - m_{s}gx - 2.5x -2.5x -0.75=0$
$-left(5+m_{s}gright)x+0.3 mg -0.75 = 0$
$$frac{0.3 mg -0.75}{5+m_{s}g}=x$$
Which in the end reproduces the same result obtained earlier. Although I don't end up with a contradiction. How am I supposed to obtain the length?. Could it be that did I overlooked something, was it my system of equations wrong? or just something in the analysis that is incomplete?. I hope somebody could guide me with this problem as I'm still stuck on it.
algebra-precalculus systems-of-equations physics
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
I've been going on circles with this problem these days, and although the problem doesn't seem to be that complicated I'm still lost at where is that missing statement or clue to solve this riddle.
The problem is as follows:
In Figure shown below, the meterstick's mass is $textrm{0.160 kg}$
and the string tension is $textrm{2.50 N}$. The system is in
equilibrium. Find (a)the unknown mass $m$ and (b) the upward force the
fulcrum exerts on the stick.
Now what I did in my attempt to solve this problem was to use this fact from Rotational equilibrium:
$$tau_{net}=0$$
Therefore to do so I had to establish the lengths of the weights attached to the meterstick, provided that there are four forces acting.
To which I used the following notation $m_{s}$ for the mass of the meterstick, $m$ mass of the unknown object, $g$ for gravity and $x$ for the unknown length to the center of mass.
From static equilibrium in $textrm{y-axis}$:
$$F_net=0$$
$$F_{1}=textrm{normal force}=mg+m_{s}g+2.5$$
With this established all was to apply the definition given earlier to rotational equilibrium first at the fulcrum:
$tau_{1}= +mgleft(0.3right)$
$tau_{2}= -m_sgleft(xright)$
$tau_{3}= -2.5left(2x+0.30 right)$
Summing all these three and equating to zero I obtained:
$0 = +mgleft(0.3right)-m_sgleft(xright)-2.5left(2x+0.30 right)$
$0.3mg-m_{s}gx-5x-0.75=0$
$$x=frac{0.3mg-0.75}{m_{s}g+5}$$
Then I thought to use the above equation combined with the one obtained from static equilibrium in the y-axis and a new analysis of the rotational equilibrium but this time in the center of mass to relate those and obtain the length in terms of the mass with the hopes of solving the system of equation.
This is shown in the diagram below:
As mentioned earlier the distance to the center of mass:
$tau_{1}= + mg left( 0.3 + xright)$
$tau_{2}= - left(mg+m_{s}g+2.5right)x$
$tau_{3}= - 2.5 left( x+0.3 right)$
By summing these up and equating to zero this becomes into:
$+ mg left( 0.3 + xright) - left(mg+m_{s}g+2.5right)x - 2.5 left( x+0.3 right) = 0$
$0.3 mg + mgx - mgx - m_{s}gx - 2.5x -2.5x -0.75=0$
$-left(5+m_{s}gright)x+0.3 mg -0.75 = 0$
$$frac{0.3 mg -0.75}{5+m_{s}g}=x$$
Which in the end reproduces the same result obtained earlier. Although I don't end up with a contradiction. How am I supposed to obtain the length?. Could it be that did I overlooked something, was it my system of equations wrong? or just something in the analysis that is incomplete?. I hope somebody could guide me with this problem as I'm still stuck on it.
algebra-precalculus systems-of-equations physics
algebra-precalculus systems-of-equations physics
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
asked Jan 20 at 10:14
Chris Steinbeck BellChris Steinbeck Bell
820315
820315
3
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
2
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
1
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26
|
show 5 more comments
3
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
2
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
1
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26
3
3
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
2
2
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
1
1
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Apparently this was a case of "overlooking" what it was obvious. In the problem it was mentioned a meterstick and this was key to solve the problem.
Just in case I'd like to quote what Merriam Webster defines this device:
a measuring stick one meter long that is marked off in centimeters and usually millimeters.
Of course, in international units it is expressed as a tool of one meter, but in the US typically it is one yard (which on that case would be better stated as yardstick). But since the units used were in meters the alluded length was the first entry.
By inserting this important piece of missing information the rest came down like a domino pieces.
The drawing from below illustrates the changes made:
Then all was left to do is to pick up from where I left: (Reminding that the analysis corresponds to the rotational equilibrium at the fulcrum as indicated in the picture from below)
The length $x$ corresponds as:
$0.3+x=0.5$
$x=0.5-0.3=0.2$
Therefore:
$x=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{2}{10}=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{1}{5}=frac{0.3mg-0.75}{m_{s}g+5}$
$m_{s}g+5=5 times 0.3mg-0.75$
$m_{s}g+5=1.5mg-3.75$
$m=frac{m_{s}g+5+3.75}{1.5g}$
Inserting the values known $g = 9.8 frac{m}{s^{2}}$ and $m_{s}= 0.16,kg$
$m=frac{0.16 times 9.8+5+3.75}{1.5 times 9.8}= frac{1.568+8.75}{14.70}=frac{10.318}{14.70}$
$m approx 0.7019,kg$
Which corresponds to what my book says is the answer for part a.
The same result can be obtained by going in the other route as suggested in the drawing from below:
Which ultimately reduces to:
$+ mg left( 0.3 + 0.2right) - left(mg+m_{s}g+2.5right)0.2 - 2.5 left( 0.2+0.3 right) = 0$
$0.5 mg - 0.2mg-0.2m_{s}g-0.5 - 1.25 = 0$
$0.3 mg = 0.2 m_{s}g+1.75$
$m= frac{0.2 m_{s}g+1.75}{0.3g} = frac{0.2 times 0.16 times 9.8 + 1.75}{2.94}$
$m approx 0.7019,kg$
Finally part b of the problem can be easily obtained from the equation established earlier:
$F= mg + m_{s}g + 2.5 = 0.7019 times 9.8 + 0.16 times 9.8 + 2.5 = 6.87862 + 1.568 + 2.5 = 10.94662,N$
Which my book reports the answer to be $10.9,N$ so it must be okay.
I'm letting this solution so anyone in my situation can benefit from looking the statements and compare that it may help you in your computations.
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
add a comment |
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$begingroup$
Apparently this was a case of "overlooking" what it was obvious. In the problem it was mentioned a meterstick and this was key to solve the problem.
Just in case I'd like to quote what Merriam Webster defines this device:
a measuring stick one meter long that is marked off in centimeters and usually millimeters.
Of course, in international units it is expressed as a tool of one meter, but in the US typically it is one yard (which on that case would be better stated as yardstick). But since the units used were in meters the alluded length was the first entry.
By inserting this important piece of missing information the rest came down like a domino pieces.
The drawing from below illustrates the changes made:
Then all was left to do is to pick up from where I left: (Reminding that the analysis corresponds to the rotational equilibrium at the fulcrum as indicated in the picture from below)
The length $x$ corresponds as:
$0.3+x=0.5$
$x=0.5-0.3=0.2$
Therefore:
$x=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{2}{10}=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{1}{5}=frac{0.3mg-0.75}{m_{s}g+5}$
$m_{s}g+5=5 times 0.3mg-0.75$
$m_{s}g+5=1.5mg-3.75$
$m=frac{m_{s}g+5+3.75}{1.5g}$
Inserting the values known $g = 9.8 frac{m}{s^{2}}$ and $m_{s}= 0.16,kg$
$m=frac{0.16 times 9.8+5+3.75}{1.5 times 9.8}= frac{1.568+8.75}{14.70}=frac{10.318}{14.70}$
$m approx 0.7019,kg$
Which corresponds to what my book says is the answer for part a.
The same result can be obtained by going in the other route as suggested in the drawing from below:
Which ultimately reduces to:
$+ mg left( 0.3 + 0.2right) - left(mg+m_{s}g+2.5right)0.2 - 2.5 left( 0.2+0.3 right) = 0$
$0.5 mg - 0.2mg-0.2m_{s}g-0.5 - 1.25 = 0$
$0.3 mg = 0.2 m_{s}g+1.75$
$m= frac{0.2 m_{s}g+1.75}{0.3g} = frac{0.2 times 0.16 times 9.8 + 1.75}{2.94}$
$m approx 0.7019,kg$
Finally part b of the problem can be easily obtained from the equation established earlier:
$F= mg + m_{s}g + 2.5 = 0.7019 times 9.8 + 0.16 times 9.8 + 2.5 = 6.87862 + 1.568 + 2.5 = 10.94662,N$
Which my book reports the answer to be $10.9,N$ so it must be okay.
I'm letting this solution so anyone in my situation can benefit from looking the statements and compare that it may help you in your computations.
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
add a comment |
$begingroup$
Apparently this was a case of "overlooking" what it was obvious. In the problem it was mentioned a meterstick and this was key to solve the problem.
Just in case I'd like to quote what Merriam Webster defines this device:
a measuring stick one meter long that is marked off in centimeters and usually millimeters.
Of course, in international units it is expressed as a tool of one meter, but in the US typically it is one yard (which on that case would be better stated as yardstick). But since the units used were in meters the alluded length was the first entry.
By inserting this important piece of missing information the rest came down like a domino pieces.
The drawing from below illustrates the changes made:
Then all was left to do is to pick up from where I left: (Reminding that the analysis corresponds to the rotational equilibrium at the fulcrum as indicated in the picture from below)
The length $x$ corresponds as:
$0.3+x=0.5$
$x=0.5-0.3=0.2$
Therefore:
$x=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{2}{10}=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{1}{5}=frac{0.3mg-0.75}{m_{s}g+5}$
$m_{s}g+5=5 times 0.3mg-0.75$
$m_{s}g+5=1.5mg-3.75$
$m=frac{m_{s}g+5+3.75}{1.5g}$
Inserting the values known $g = 9.8 frac{m}{s^{2}}$ and $m_{s}= 0.16,kg$
$m=frac{0.16 times 9.8+5+3.75}{1.5 times 9.8}= frac{1.568+8.75}{14.70}=frac{10.318}{14.70}$
$m approx 0.7019,kg$
Which corresponds to what my book says is the answer for part a.
The same result can be obtained by going in the other route as suggested in the drawing from below:
Which ultimately reduces to:
$+ mg left( 0.3 + 0.2right) - left(mg+m_{s}g+2.5right)0.2 - 2.5 left( 0.2+0.3 right) = 0$
$0.5 mg - 0.2mg-0.2m_{s}g-0.5 - 1.25 = 0$
$0.3 mg = 0.2 m_{s}g+1.75$
$m= frac{0.2 m_{s}g+1.75}{0.3g} = frac{0.2 times 0.16 times 9.8 + 1.75}{2.94}$
$m approx 0.7019,kg$
Finally part b of the problem can be easily obtained from the equation established earlier:
$F= mg + m_{s}g + 2.5 = 0.7019 times 9.8 + 0.16 times 9.8 + 2.5 = 6.87862 + 1.568 + 2.5 = 10.94662,N$
Which my book reports the answer to be $10.9,N$ so it must be okay.
I'm letting this solution so anyone in my situation can benefit from looking the statements and compare that it may help you in your computations.
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
add a comment |
$begingroup$
Apparently this was a case of "overlooking" what it was obvious. In the problem it was mentioned a meterstick and this was key to solve the problem.
Just in case I'd like to quote what Merriam Webster defines this device:
a measuring stick one meter long that is marked off in centimeters and usually millimeters.
Of course, in international units it is expressed as a tool of one meter, but in the US typically it is one yard (which on that case would be better stated as yardstick). But since the units used were in meters the alluded length was the first entry.
By inserting this important piece of missing information the rest came down like a domino pieces.
The drawing from below illustrates the changes made:
Then all was left to do is to pick up from where I left: (Reminding that the analysis corresponds to the rotational equilibrium at the fulcrum as indicated in the picture from below)
The length $x$ corresponds as:
$0.3+x=0.5$
$x=0.5-0.3=0.2$
Therefore:
$x=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{2}{10}=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{1}{5}=frac{0.3mg-0.75}{m_{s}g+5}$
$m_{s}g+5=5 times 0.3mg-0.75$
$m_{s}g+5=1.5mg-3.75$
$m=frac{m_{s}g+5+3.75}{1.5g}$
Inserting the values known $g = 9.8 frac{m}{s^{2}}$ and $m_{s}= 0.16,kg$
$m=frac{0.16 times 9.8+5+3.75}{1.5 times 9.8}= frac{1.568+8.75}{14.70}=frac{10.318}{14.70}$
$m approx 0.7019,kg$
Which corresponds to what my book says is the answer for part a.
The same result can be obtained by going in the other route as suggested in the drawing from below:
Which ultimately reduces to:
$+ mg left( 0.3 + 0.2right) - left(mg+m_{s}g+2.5right)0.2 - 2.5 left( 0.2+0.3 right) = 0$
$0.5 mg - 0.2mg-0.2m_{s}g-0.5 - 1.25 = 0$
$0.3 mg = 0.2 m_{s}g+1.75$
$m= frac{0.2 m_{s}g+1.75}{0.3g} = frac{0.2 times 0.16 times 9.8 + 1.75}{2.94}$
$m approx 0.7019,kg$
Finally part b of the problem can be easily obtained from the equation established earlier:
$F= mg + m_{s}g + 2.5 = 0.7019 times 9.8 + 0.16 times 9.8 + 2.5 = 6.87862 + 1.568 + 2.5 = 10.94662,N$
Which my book reports the answer to be $10.9,N$ so it must be okay.
I'm letting this solution so anyone in my situation can benefit from looking the statements and compare that it may help you in your computations.
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
$endgroup$
Apparently this was a case of "overlooking" what it was obvious. In the problem it was mentioned a meterstick and this was key to solve the problem.
Just in case I'd like to quote what Merriam Webster defines this device:
a measuring stick one meter long that is marked off in centimeters and usually millimeters.
Of course, in international units it is expressed as a tool of one meter, but in the US typically it is one yard (which on that case would be better stated as yardstick). But since the units used were in meters the alluded length was the first entry.
By inserting this important piece of missing information the rest came down like a domino pieces.
The drawing from below illustrates the changes made:
Then all was left to do is to pick up from where I left: (Reminding that the analysis corresponds to the rotational equilibrium at the fulcrum as indicated in the picture from below)
The length $x$ corresponds as:
$0.3+x=0.5$
$x=0.5-0.3=0.2$
Therefore:
$x=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{2}{10}=frac{0.3mg-0.75}{m_{s}g+5}$
$frac{1}{5}=frac{0.3mg-0.75}{m_{s}g+5}$
$m_{s}g+5=5 times 0.3mg-0.75$
$m_{s}g+5=1.5mg-3.75$
$m=frac{m_{s}g+5+3.75}{1.5g}$
Inserting the values known $g = 9.8 frac{m}{s^{2}}$ and $m_{s}= 0.16,kg$
$m=frac{0.16 times 9.8+5+3.75}{1.5 times 9.8}= frac{1.568+8.75}{14.70}=frac{10.318}{14.70}$
$m approx 0.7019,kg$
Which corresponds to what my book says is the answer for part a.
The same result can be obtained by going in the other route as suggested in the drawing from below:
Which ultimately reduces to:
$+ mg left( 0.3 + 0.2right) - left(mg+m_{s}g+2.5right)0.2 - 2.5 left( 0.2+0.3 right) = 0$
$0.5 mg - 0.2mg-0.2m_{s}g-0.5 - 1.25 = 0$
$0.3 mg = 0.2 m_{s}g+1.75$
$m= frac{0.2 m_{s}g+1.75}{0.3g} = frac{0.2 times 0.16 times 9.8 + 1.75}{2.94}$
$m approx 0.7019,kg$
Finally part b of the problem can be easily obtained from the equation established earlier:
$F= mg + m_{s}g + 2.5 = 0.7019 times 9.8 + 0.16 times 9.8 + 2.5 = 6.87862 + 1.568 + 2.5 = 10.94662,N$
Which my book reports the answer to be $10.9,N$ so it must be okay.
I'm letting this solution so anyone in my situation can benefit from looking the statements and compare that it may help you in your computations.
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
answered Jan 20 at 17:22
Chris Steinbeck BellChris Steinbeck Bell
820315
820315
add a comment |
add a comment |
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3
$begingroup$
Doesn't this fit better on Physics SE?
$endgroup$
– Zacky
Jan 20 at 10:18
$begingroup$
Physics SE stopped taking "homework" questions a long while ago.
$endgroup$
– Raskolnikov
Jan 20 at 10:20
$begingroup$
@Zacky As one commenter mentioned by the nature how this question is stated probably would be off topic.
$endgroup$
– Chris Steinbeck Bell
Jan 20 at 10:22
2
$begingroup$
@ChrisSteinbeckBell By definition of the word, a meterstick has length 1 meter. Although I can see how one might take it to mean a stick to measure (various) meters.
$endgroup$
– Daniel Pietrobon
Jan 20 at 10:34
1
$begingroup$
Clearly x = .2 m. What is the conversion from N to g?
$endgroup$
– William Elliot
Jan 20 at 11:26