Verbundene Massen
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
Need help? Yes, please!
The following quantities appear in the problem:
The following formulas must be used to solve the exercise:
No explanation / solution video for this exercise has yet been created.
But there is a video to a similar exercise:
In case your browser prevents YouTube embedding: https://youtu.be/rbJpXjt7DZk
But there is a video to a similar exercise:
Exercise:
Eine Stahlkiste der Masse .kg befindet sich auf einer schiefen Stahlebene mit Neigungswinkel degree. Durch ein Seil das über eine Rolle geführt wird ist sie mit einem Stein verbunden. Welche Masse darf der Stein maximal haben damit die Anordnung gerade noch in Ruhe ist? Die Seilmasse soll vernachlässigt werden. center tikzpicturescale force/.stylelatexdrawbluefillblue axis/.styledensely dashedgrayfontsmall M/.stylerectangledrawfilllightgrayminimum size.cmthin m/.stylerectangledrawblackfilllightgrayminimum size.cmthin plane/.styledrawblackfillblue! string/.styledrawred thick pulley/.stylethick defiangle % Angle of the inclined plane defdown- defarcr.cm % Radius of the arc used to indicate angles drawplane - coordinate base -- coordinatepos. mid ++iangle: coordinate top |- base -- cycle; path mid nodeMrotateiangleyshift.cm M ; drawpulley top -- ++iangle:. circle .cm++ -iangle:. coordinate pulley; drawstring M.east -- ++iangle:.cm arc +iangle::.-- ++- nodem ; draw- base++arcr arc :iangle:arcr; path base++iangle*.:arcr+pt node alpha; tikzpicture center
Solution:
newqtyM.kg newqtyadegree newnumgl. % Geg M M alpha a % GesMassem sikg % Die Hangabtriebskraft beträgt solqtyFpMgsinalphaMn*ncgn*sindanN al Fpara Fpf M ncg sina Fp % die Reibungskraft solqtyFrsscmuHMgcosalphagln*Mn*ncgn*cosdanN al sscFR Frf gl M ncg cosa Fr. Die Gewichtskraft des Steins muss folglich solqtyFgMgsinalpha + sscmuHMgcosalphaFpn+FrnN al sscFG Fpara + FR Fgf Fp - Fr Fg sein. Somit ist seine Masse solqtymMsinalpha + sscmuHcosalphaFgn/ncgnkg al m fracFGg mf fracFgncg m. % m mf mTT
Eine Stahlkiste der Masse .kg befindet sich auf einer schiefen Stahlebene mit Neigungswinkel degree. Durch ein Seil das über eine Rolle geführt wird ist sie mit einem Stein verbunden. Welche Masse darf der Stein maximal haben damit die Anordnung gerade noch in Ruhe ist? Die Seilmasse soll vernachlässigt werden. center tikzpicturescale force/.stylelatexdrawbluefillblue axis/.styledensely dashedgrayfontsmall M/.stylerectangledrawfilllightgrayminimum size.cmthin m/.stylerectangledrawblackfilllightgrayminimum size.cmthin plane/.styledrawblackfillblue! string/.styledrawred thick pulley/.stylethick defiangle % Angle of the inclined plane defdown- defarcr.cm % Radius of the arc used to indicate angles drawplane - coordinate base -- coordinatepos. mid ++iangle: coordinate top |- base -- cycle; path mid nodeMrotateiangleyshift.cm M ; drawpulley top -- ++iangle:. circle .cm++ -iangle:. coordinate pulley; drawstring M.east -- ++iangle:.cm arc +iangle::.-- ++- nodem ; draw- base++arcr arc :iangle:arcr; path base++iangle*.:arcr+pt node alpha; tikzpicture center
Solution:
newqtyM.kg newqtyadegree newnumgl. % Geg M M alpha a % GesMassem sikg % Die Hangabtriebskraft beträgt solqtyFpMgsinalphaMn*ncgn*sindanN al Fpara Fpf M ncg sina Fp % die Reibungskraft solqtyFrsscmuHMgcosalphagln*Mn*ncgn*cosdanN al sscFR Frf gl M ncg cosa Fr. Die Gewichtskraft des Steins muss folglich solqtyFgMgsinalpha + sscmuHMgcosalphaFpn+FrnN al sscFG Fpara + FR Fgf Fp - Fr Fg sein. Somit ist seine Masse solqtymMsinalpha + sscmuHcosalphaFgn/ncgnkg al m fracFGg mf fracFgncg m. % m mf mTT
Meta Information
Exercise:
Eine Stahlkiste der Masse .kg befindet sich auf einer schiefen Stahlebene mit Neigungswinkel degree. Durch ein Seil das über eine Rolle geführt wird ist sie mit einem Stein verbunden. Welche Masse darf der Stein maximal haben damit die Anordnung gerade noch in Ruhe ist? Die Seilmasse soll vernachlässigt werden. center tikzpicturescale force/.stylelatexdrawbluefillblue axis/.styledensely dashedgrayfontsmall M/.stylerectangledrawfilllightgrayminimum size.cmthin m/.stylerectangledrawblackfilllightgrayminimum size.cmthin plane/.styledrawblackfillblue! string/.styledrawred thick pulley/.stylethick defiangle % Angle of the inclined plane defdown- defarcr.cm % Radius of the arc used to indicate angles drawplane - coordinate base -- coordinatepos. mid ++iangle: coordinate top |- base -- cycle; path mid nodeMrotateiangleyshift.cm M ; drawpulley top -- ++iangle:. circle .cm++ -iangle:. coordinate pulley; drawstring M.east -- ++iangle:.cm arc +iangle::.-- ++- nodem ; draw- base++arcr arc :iangle:arcr; path base++iangle*.:arcr+pt node alpha; tikzpicture center
Solution:
newqtyM.kg newqtyadegree newnumgl. % Geg M M alpha a % GesMassem sikg % Die Hangabtriebskraft beträgt solqtyFpMgsinalphaMn*ncgn*sindanN al Fpara Fpf M ncg sina Fp % die Reibungskraft solqtyFrsscmuHMgcosalphagln*Mn*ncgn*cosdanN al sscFR Frf gl M ncg cosa Fr. Die Gewichtskraft des Steins muss folglich solqtyFgMgsinalpha + sscmuHMgcosalphaFpn+FrnN al sscFG Fpara + FR Fgf Fp - Fr Fg sein. Somit ist seine Masse solqtymMsinalpha + sscmuHcosalphaFgn/ncgnkg al m fracFGg mf fracFgncg m. % m mf mTT
Eine Stahlkiste der Masse .kg befindet sich auf einer schiefen Stahlebene mit Neigungswinkel degree. Durch ein Seil das über eine Rolle geführt wird ist sie mit einem Stein verbunden. Welche Masse darf der Stein maximal haben damit die Anordnung gerade noch in Ruhe ist? Die Seilmasse soll vernachlässigt werden. center tikzpicturescale force/.stylelatexdrawbluefillblue axis/.styledensely dashedgrayfontsmall M/.stylerectangledrawfilllightgrayminimum size.cmthin m/.stylerectangledrawblackfilllightgrayminimum size.cmthin plane/.styledrawblackfillblue! string/.styledrawred thick pulley/.stylethick defiangle % Angle of the inclined plane defdown- defarcr.cm % Radius of the arc used to indicate angles drawplane - coordinate base -- coordinatepos. mid ++iangle: coordinate top |- base -- cycle; path mid nodeMrotateiangleyshift.cm M ; drawpulley top -- ++iangle:. circle .cm++ -iangle:. coordinate pulley; drawstring M.east -- ++iangle:.cm arc +iangle::.-- ++- nodem ; draw- base++arcr arc :iangle:arcr; path base++iangle*.:arcr+pt node alpha; tikzpicture center
Solution:
newqtyM.kg newqtyadegree newnumgl. % Geg M M alpha a % GesMassem sikg % Die Hangabtriebskraft beträgt solqtyFpMgsinalphaMn*ncgn*sindanN al Fpara Fpf M ncg sina Fp % die Reibungskraft solqtyFrsscmuHMgcosalphagln*Mn*ncgn*cosdanN al sscFR Frf gl M ncg cosa Fr. Die Gewichtskraft des Steins muss folglich solqtyFgMgsinalpha + sscmuHMgcosalphaFpn+FrnN al sscFG Fpara + FR Fgf Fp - Fr Fg sein. Somit ist seine Masse solqtymMsinalpha + sscmuHcosalphaFgn/ncgnkg al m fracFGg mf fracFgncg m. % m mf mTT
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Masse auf schiefer Ebene by TeXercises
This is the original exercise.
Title | Creator | |||
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Verbundene Massen | aej |