From 3627259a2b60564f60fb3979090a6dc03ee603b2 Mon Sep 17 00:00:00 2001 From: brynnchernosky <56202540+brynnchernosky@users.noreply.github.com> Date: Mon, 1 May 2023 17:37:43 -0400 Subject: add json files --- .../PhysicsBox/PhysicsSimulationQuestions.json | 161 +++++++++++++++++++++ 1 file changed, 161 insertions(+) create mode 100644 src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json (limited to 'src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json') diff --git a/src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json b/src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json new file mode 100644 index 000000000..cc79f7aad --- /dev/null +++ b/src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json @@ -0,0 +1,161 @@ +{ + "inclinePlane": [ + { + "questionSetup": [ + "There is a 1kg weight on an inclined plane. The plane is at a ", + " angle from the ground. The system is in equilibrium (the net force on the weight is 0)." + ], + "variablesForQuestionSetup": ["theta - max 45"], + "question": "What are the magnitudes and directions of the forces acting on the weight?", + "answerParts": [ + "force of gravity", + "angle of gravity", + "normal force", + "angle of normal force", + "force of static friction", + "angle of static friction" + ], + "answerSolutionDescriptions": [ + "9.81", + "270", + "solve normal force magnitude from wedge angle", + "solve normal force angle from wedge angle", + "solve static force magnitude from wedge angle given equilibrium", + "solve static force angle from wedge angle given equilibrium" + ], + "goal": "noMovement", + "hints": [ + { + "description": "Direction of Force of Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad." + }, + { + "description": "Direction of Normal Force", + "content": "The normal force acts in the direction perpendicular to the incline plane: π/2-θ rad, where θ is the angle of the incline plane." + }, + { + "description": "Direction of Force of Friction", + "content": "The force of friction acts in the direction along the incline plane: π-θ rad, where θ is the angle of the incline plane." + }, + { + "description": "Magnitude of Force of Gravity", + "content": "The magnitude of the force of gravity is approximately 9.81." + }, + { + "description": "Magnitude of Normal Force", + "content": "The magnitude of the normal force is equal to m*g*cos(θ), where θ is the angle of the incline plane." + }, + { + "description": "Net Force in Equilibrium", + "content": "For the system to be in equilibrium, the sum of the x components of all forces must equal 0, and the sum of the y components of all forces must equal 0." + }, + { + "description": "X Component of Normal Force", + "content": "The x component of the normal force is equal to m*g*cos(θ)*cos(π/2-θ), where θ is the angle of the incline plane." + }, + { + "description": "X Component of Force of Friction", + "content": "Since the net force in the x direction must be 0, we know the magnitude of the x component of the friction force is m*g*cos(θ)*cos(π/2-θ)." + }, + { + "description": "Y Component of Normal Force", + "content": "The y component of the normal force is equal to m*g*cos(θ)*sin(π/2-θ), where θ is the angle of the incline plane. The y component of gravity is equal to m*g" + }, + { + "description": "Y Component of Force of Friction", + "content": "Since the net force in the x direction must be 0, we know the magnitude of the y component of the friction force is m*g-m*g*cos(θ)*sin(π/2-θ)." + }, + { + "description": "Magnitude of Force of Friction", + "content": "Combining the x and y components of the friction force, we get the magnitude of the friction force is equal to sqrt((m*g*cos(θ)*cos(π/2-θ))^2 + (m*g-m*g*cos(θ)*sin(π/2-θ))^2)." + } + ] + }, + { + "questionSetup": [ + "There is a 1kg weight on an inclined plane. The plane is at a ", + " angle from the ground. The system is in equilibrium (the net force on the weight is 0)." + ], + "variablesForQuestionSetup": ["theta - max 45"], + "question": "What is the minimum coefficient of static friction?", + "answerParts": ["coefficient of static friction"], + "answerSolutionDescriptions": [ + "solve minimum static coefficient from wedge angle given equilibrium" + ], + "goal": "noMovement", + "hints": [ + { + "description": "Net Force in Equilibrium", + "content": "If the system is in equilibrium, the sum of the x components of all forces must equal 0. In this system, the normal force and force of static friction have non-zero x components." + }, + { + "description": "X Component of Normal Force", + "content": "The x component of the normal force is equal to m*g*cos(θ)*cos(π/2-θ), where θ is the angle of the incline plane." + }, + { + "description": "X Component of Force of Friction", + "content": "The x component of the force of static friction is equal to μ*m*g*cos(θ)*cos(π-θ), where θ is the angle of the incline plane." + }, + { + "description": "Equation to Solve for Minimum Coefficient of Static Friction", + "content": "Since the net force in the x direction must be 0, we can solve the equation 0=m*g*cos(θ)*cos(π/2-θ)+μ*m*g*cos(θ)*cos(π-θ) for μ to find the minimum coefficient of static friction such that the system stays in equilibrium." + } + ] + }, + { + "questionSetup": [ + "There is a 1kg weight on an inclined plane. The coefficient of static friction is ", + ". The system is in equilibrium (the net force on the weight is 0)." + ], + "variablesForQuestionSetup": ["coefficient of static friction"], + "question": "What is the maximum angle of the plane from the ground?", + "answerParts": ["wedge angle"], + "answerSolutionDescriptions": [ + "solve maximum wedge angle from coefficient of static friction given equilibrium" + ], + "goal": "noMovement", + "hints": [ + { + "description": "Net Force in Equilibrium", + "content": "If the system is in equilibrium, the sum of the x components of all forces must equal 0. In this system, the normal force and force of static friction have non-zero x components." + }, + { + "description": "X Component of Normal Force", + "content": "The x component of the normal force is equal to m*g*cos(θ)*cos(π/2-θ), where θ is the angle of the incline plane." + }, + { + "description": "X Component of Force of Friction", + "content": "The x component of the force of static friction is equal to μ*m*g*cos(θ)*cos(π-θ), where θ is the angle of the incline plane." + }, + { + "description": "Equation to Solve for Maximum Wedge Angle", + "content": "Since the net force in the x direction must be 0, we can solve the equation 0=m*g*cos(θ)*cos(π/2-θ)+μ*m*g*cos(θ)*cos(π-θ) for θ to find the maximum wedge angle such that the system stays in equilibrium." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "Simplifying 0=m*g*cos(θ)*cos(π/2-θ)+μ*m*g*cos(θ)*cos(π-θ), we get cos(π/2-θ)=-μ*cos(π-θ)." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "The cosine subtraction formula states that cos(A-B)=cos(A)*cos(B)+sin(A)sin(B)." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "Applying the cosine subtraction formula to cos(π/2-θ)=-μ*cos(π-θ), we get cos(π/2)*cos(θ)+sin(π/2)*sin(θ)=-μ*(cos(π)cos(θ)+sin(π)sin(θ))." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "Simplifying cos(π/2)*cos(θ)-sin(π/2)*sin(θ)=-μ*(cos(π)cos(θ)-sin(π)sin(θ)), we get -sin(θ)=-μ*(-cos(θ))." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "Simplifying -sin(θ)=-μ*(-cos(θ)), we get tan(θ)=-μ." + }, + { + "description": "Simplifying Equation to Solve for Maximum Wedge Angle", + "content": "Solving for θ, we get θ = atan(μ)." + } + ] + } + ] +} -- cgit v1.2.3-70-g09d2