Wirtshausschild
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:
Kraft \(F\) / Winkel \(\theta\) /
The following formulas must be used to solve the exercise:
\(F_x = F \cos\alpha \quad F_y = F \sin\alpha \quad \) \(\sum \vec F \stackrel{!}{=} \vec 0 \quad \)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
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Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Eine Krone hängt als Wirtshausschild an der skizzierten Stabverbindung. Der obere Stab übe .kN der untere daN auf den Aufhängepunkt P aus. Die Winkel sind der Skizze zu entnehmen. abclist abc Zeichne die Gewichtskraft der Krone ein. abc Zeichne die im Text genannten Kräfte in der Skizze ein zerlege sie sinnvoll in Komponenten. abc Berechne die Gewichtskraft der Krone. abclist Schreibe alle Kraftvektoren sinnvoll an. Unbeschriftete Kraftvektoren geben keine Punkte! center tikzpicturescale. filldrawdrawblack!!white fillblack!!white ---.---.------cycle; drawcolorblack thick ---.; drawcolorblack thick --; drawcolorblack thick ---; drawcolorblack thick --++-; fill circle pt nodeabove rightP; drawthick -- ++-.-- ++--++.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.-- cycle; draw . nodeleft xshift-.cm yshift.cmdegree arc .::.cm; draw nodeleft xshift-.cmyshift-.cmdegree arc ::.cm; tikzpicture center
Solution:
NewQtyAdegree NewQtyBdegree NewQtyFo.kNeN NewQtyFudaNeN % Geg sscFo FoO Fo sscFu FuO Fu alpha A beta B % GesGewichtskraftsscFG siN % Die Vertikalkomponente der Kraft des oberen Stabes beträgt SolQtyFovsscFocosalphaFoX*cosdAXN al sscFov FovF FocosA Fov diejenige der Kraft des unteren Stabes SolQtyFuvsscFucosbetaFuX*cosdBXN al sscFuv FuvF FucosB Fuv. Die Kraft des oberen Stabes wirkt in die Richtung der Gewichtskraft diejenige das unteren Stabes gegen die Richtung der Gewichtskraft. Die Gewichtskraft ist deshalb die Differenz der Vertikalkomponenten des unteren und des oberen Stabes: SolQtyFGFuvF-FovFFuvX-FovXN al sscFG sscFuv-sscFov FGF Fuv-Fov FG % sscFG FGF FGS
Eine Krone hängt als Wirtshausschild an der skizzierten Stabverbindung. Der obere Stab übe .kN der untere daN auf den Aufhängepunkt P aus. Die Winkel sind der Skizze zu entnehmen. abclist abc Zeichne die Gewichtskraft der Krone ein. abc Zeichne die im Text genannten Kräfte in der Skizze ein zerlege sie sinnvoll in Komponenten. abc Berechne die Gewichtskraft der Krone. abclist Schreibe alle Kraftvektoren sinnvoll an. Unbeschriftete Kraftvektoren geben keine Punkte! center tikzpicturescale. filldrawdrawblack!!white fillblack!!white ---.---.------cycle; drawcolorblack thick ---.; drawcolorblack thick --; drawcolorblack thick ---; drawcolorblack thick --++-; fill circle pt nodeabove rightP; drawthick -- ++-.-- ++--++.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.-- cycle; draw . nodeleft xshift-.cm yshift.cmdegree arc .::.cm; draw nodeleft xshift-.cmyshift-.cmdegree arc ::.cm; tikzpicture center
Solution:
NewQtyAdegree NewQtyBdegree NewQtyFo.kNeN NewQtyFudaNeN % Geg sscFo FoO Fo sscFu FuO Fu alpha A beta B % GesGewichtskraftsscFG siN % Die Vertikalkomponente der Kraft des oberen Stabes beträgt SolQtyFovsscFocosalphaFoX*cosdAXN al sscFov FovF FocosA Fov diejenige der Kraft des unteren Stabes SolQtyFuvsscFucosbetaFuX*cosdBXN al sscFuv FuvF FucosB Fuv. Die Kraft des oberen Stabes wirkt in die Richtung der Gewichtskraft diejenige das unteren Stabes gegen die Richtung der Gewichtskraft. Die Gewichtskraft ist deshalb die Differenz der Vertikalkomponenten des unteren und des oberen Stabes: SolQtyFGFuvF-FovFFuvX-FovXN al sscFG sscFuv-sscFov FGF Fuv-Fov FG % sscFG FGF FGS
Meta Information
Exercise:
Eine Krone hängt als Wirtshausschild an der skizzierten Stabverbindung. Der obere Stab übe .kN der untere daN auf den Aufhängepunkt P aus. Die Winkel sind der Skizze zu entnehmen. abclist abc Zeichne die Gewichtskraft der Krone ein. abc Zeichne die im Text genannten Kräfte in der Skizze ein zerlege sie sinnvoll in Komponenten. abc Berechne die Gewichtskraft der Krone. abclist Schreibe alle Kraftvektoren sinnvoll an. Unbeschriftete Kraftvektoren geben keine Punkte! center tikzpicturescale. filldrawdrawblack!!white fillblack!!white ---.---.------cycle; drawcolorblack thick ---.; drawcolorblack thick --; drawcolorblack thick ---; drawcolorblack thick --++-; fill circle pt nodeabove rightP; drawthick -- ++-.-- ++--++.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.-- cycle; draw . nodeleft xshift-.cm yshift.cmdegree arc .::.cm; draw nodeleft xshift-.cmyshift-.cmdegree arc ::.cm; tikzpicture center
Solution:
NewQtyAdegree NewQtyBdegree NewQtyFo.kNeN NewQtyFudaNeN % Geg sscFo FoO Fo sscFu FuO Fu alpha A beta B % GesGewichtskraftsscFG siN % Die Vertikalkomponente der Kraft des oberen Stabes beträgt SolQtyFovsscFocosalphaFoX*cosdAXN al sscFov FovF FocosA Fov diejenige der Kraft des unteren Stabes SolQtyFuvsscFucosbetaFuX*cosdBXN al sscFuv FuvF FucosB Fuv. Die Kraft des oberen Stabes wirkt in die Richtung der Gewichtskraft diejenige das unteren Stabes gegen die Richtung der Gewichtskraft. Die Gewichtskraft ist deshalb die Differenz der Vertikalkomponenten des unteren und des oberen Stabes: SolQtyFGFuvF-FovFFuvX-FovXN al sscFG sscFuv-sscFov FGF Fuv-Fov FG % sscFG FGF FGS
Eine Krone hängt als Wirtshausschild an der skizzierten Stabverbindung. Der obere Stab übe .kN der untere daN auf den Aufhängepunkt P aus. Die Winkel sind der Skizze zu entnehmen. abclist abc Zeichne die Gewichtskraft der Krone ein. abc Zeichne die im Text genannten Kräfte in der Skizze ein zerlege sie sinnvoll in Komponenten. abc Berechne die Gewichtskraft der Krone. abclist Schreibe alle Kraftvektoren sinnvoll an. Unbeschriftete Kraftvektoren geben keine Punkte! center tikzpicturescale. filldrawdrawblack!!white fillblack!!white ---.---.------cycle; drawcolorblack thick ---.; drawcolorblack thick --; drawcolorblack thick ---; drawcolorblack thick --++-; fill circle pt nodeabove rightP; drawthick -- ++-.-- ++--++.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.--++-.+.--++-.-.-- cycle; draw . nodeleft xshift-.cm yshift.cmdegree arc .::.cm; draw nodeleft xshift-.cmyshift-.cmdegree arc ::.cm; tikzpicture center
Solution:
NewQtyAdegree NewQtyBdegree NewQtyFo.kNeN NewQtyFudaNeN % Geg sscFo FoO Fo sscFu FuO Fu alpha A beta B % GesGewichtskraftsscFG siN % Die Vertikalkomponente der Kraft des oberen Stabes beträgt SolQtyFovsscFocosalphaFoX*cosdAXN al sscFov FovF FocosA Fov diejenige der Kraft des unteren Stabes SolQtyFuvsscFucosbetaFuX*cosdBXN al sscFuv FuvF FucosB Fuv. Die Kraft des oberen Stabes wirkt in die Richtung der Gewichtskraft diejenige das unteren Stabes gegen die Richtung der Gewichtskraft. Die Gewichtskraft ist deshalb die Differenz der Vertikalkomponenten des unteren und des oberen Stabes: SolQtyFGFuvF-FovFFuvX-FovXN al sscFG sscFuv-sscFov FGF Fuv-Fov FG % sscFG FGF FGS
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