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| -rw-r--r-- | src/client/views/nodes/PhysicsBox/PhysicsSimulationQuestions.json | 161 | ||||
| -rw-r--r-- | src/client/views/nodes/PhysicsBox/PhysicsSimulationTutorial.json | 472 |
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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)." + } + ] + } + ] +} diff --git a/src/client/views/nodes/PhysicsBox/PhysicsSimulationTutorial.json b/src/client/views/nodes/PhysicsBox/PhysicsSimulationTutorial.json new file mode 100644 index 000000000..16d1de96c --- /dev/null +++ b/src/client/views/nodes/PhysicsBox/PhysicsSimulationTutorial.json @@ -0,0 +1,472 @@ +{ + "freeWeight": { + "question": "A 1kg weight is at rest on the ground. What are the magnitude and directions of the forces acting on the weight?", + "steps": [ + { + "description": "Forces", + "content": "There are two forces acting on the weight: the force of gravity and the normal force.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Normal Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Force of Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "Normal Force", + "content": "The normal force acts in the positive y direction: π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Normal Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "All Forces", + "content": "Combining all of the forces, we get the following free body diagram.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Normal Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "pendulum": { + "question": "A 1kg weight on a 300m rod of negligible mass is released from an angle 30 degrees offset from equilibrium. What are the magnitude and directions of the forces acting on the weight immediately after release? (Ignore air resistance)", + "steps": [ + { + "description": "Forces", + "content": "There are two force acting on the weight: the force of gravity and the force of tension.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Tension", + "magnitude": 8.5, + "directionInDegrees": 60, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Force of Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "Tension", + "content": "The force of tension acts along the direction of the rod. The rod is 30 degrees offset from equilibrium, so the direction along the rod is 90-30=60 degrees. The tension force has two components—the component creating the centripetal force and the component canceling out the parallel component of gravity. The weight has just been released, so it has velocity 0, meaning the centripetal force is 0. Thus, the tension force only acts to cancel out the parallel component of gravity. Thus, the magnitude of tension is m*g*sin(60°)", + "forces": [ + { + "description": "Tension", + "magnitude": 8.5, + "directionInDegrees": 60, + "component": false + }, + { + "description": "Gravity - Parallel Component", + "magnitude": 8.5, + "directionInDegrees": 240, + "component": true + } + ], + "showMagnitude": true + }, + { + "description": "All Forces", + "content": "Combining all of the forces, we get the following free body diagram.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Tension", + "magnitude": 8.5, + "directionInDegrees": 60, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "inclinePlane": { + "question": "There is a 1kg weight on an inclined plane. The plane is at an angle θ from the ground, and has a coefficient of static friction μ. The system is in equilibrium (the net force on the weight is 0). What are the magnitudes and directions of the forces acting on the weight?", + "steps": [ + { + "description": "Forces", + "content": "There are three forces acting on the weight: the force of gravity, the normal force, and the force of static friction.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Normal Force", + "magnitude": 8.817, + "directionInDegrees": 64, + "component": false + }, + { + "description": "Friction Force", + "magnitude": 4.3, + "directionInDegrees": 154, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Force of Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "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. The magnitude of the normal force is equal to m*g*cos(θ).", + "forces": [ + { + "description": "Normal Force", + "magnitude": 8.817, + "directionInDegrees": 64, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "Force of Static Friction", + "content": "The force of static friction acts in the direction along the incline plane: π-θ rad, where θ is the angle of the incline plane. We can use the knowledge that the system is in equilibrium to solve for the magnitude of the force of static friction.", + "forces": [ + { + "description": "Friction Force", + "magnitude": 4.3, + "directionInDegrees": 154, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Net X Force in Equilibrium", + "content": "For the system to be in equilibrium, the sum of the x components of all forces must equal 0. 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. The x component of gravity is equal to 0. 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).", + "forces": [ + { + "description": "Normal Force - X Component", + "magnitude": 3.87, + "directionInDegrees": 0, + "component": true + }, + { + "description": "Friction Force - X Component", + "magnitude": 3.87, + "directionInDegrees": 180, + "component": true + } + ], + "showMagnitude": true + }, + { + "description": "Net Y Force Normal Force", + "content": "For the system to be in equilibrium, the sum of the y components of all forces must equal 0. 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. 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).", + "forces": [ + { + "description": "Normal Force - Y Component ", + "magnitude": 7.92, + "directionInDegrees": 90, + "component": true + }, + { + "description": "Gravity - Y Component ", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": true + }, + { + "description": "Friction Force - Y Component ", + "magnitude": 1.89, + "directionInDegrees": 90, + "component": true + } + ], + "showMagnitude": true + }, + { + "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*cos(θ)*sin(π/2-θ)-m*g)^2).", + "forces": [ + { + "description": "Friction Force", + "magnitude": 4.3, + "directionInDegrees": 154, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "All Forces", + "content": "Combining all of the forces, we get the following free body diagram.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Normal Force", + "magnitude": 8.817, + "directionInDegrees": 64, + "component": false + }, + { + "description": "Friction Force", + "magnitude": 4.3, + "directionInDegrees": 154, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "spring": { + "question": "A 1kg weight is on a spring of negligible mass with rest length 200m and spring constant 0.5. What is the equilibrium spring length?", + "steps": [ + { + "description": "Forces", + "content": "We can start by solving for the forces acting on the weight at any given point in time. There are two forces potentially acting on the weight: the force of gravity and the spring force. In equilibrium, these forces will be perfectly balanced.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Spring Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Force of Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "Spring Force", + "content": "The spring force acts in the negative y direction (3π/2 rad) if the spring is compressed. The spring force acts in the positive y direction (π/2 rad) if the spring is extended. Because the forces are perfectly balanced and gravity acts in the negative y direction, the spring force must act in the positive y direction and have the same magnitude as the force og gravity, m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Spring Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + }, + { + "description": "Spring Force", + "content": "We can use the spring force equation, Fs=kd to solve for the displacement such that Fs=mg. Setting them equal, we get mg=kd. Plugging in for the known values of m,g, and k, we get 1*9.81=0.5d. Solving for d, we get d=19.62 as the equilibrium starting displacement", + "forces": [ + { + "description": "Spring Force", + "magnitude": 9.81, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "circular": { + "question": "A 1kg weight is attached to a 100m rod of negligible mass. The weight is undergoing uniform circular motion with tangential velocity 40 m/s. What are the magnitude and directions of the forces acting on the weight? (Ignore air resistance)", + "steps": [ + { + "description": "Forces", + "content": "There is one force acting on the weight: the centripetal force.", + "forces": [ + { + "description": "Centripetal Force", + "magnitude": 16, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Centripetal Force", + "content": "The centripetal force is always directed toward the center of the circle. The formula for solving for the magnitude of centripetal force for an object undergoing uniform circular motion is Fc=mv^2 / r. Plugging in for known values, we get Fc=1*(40^2)/100. Solving for this, we get Fc=16", + "forces": [ + { + "description": "Centripetal Force", + "magnitude": 16, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "pulley": { + "question": "A 1kg red weight is attached to a simple pulley with a rope of negligible mass. A 1.5kg blue weight is attached to the other end of the simple pulley. What are the forces acting on the red weight?", + "steps": [ + { + "description": "Forces", + "content": "There are two force acting on the red weight: the force of gravity and the force of tension.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Tension", + "magnitude": 16.68, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": false + }, + { + "description": "Gravity", + "content": "The force of gravity acts in the negative y direction: 3π/2 rad. It has magnitude equal to m*g. We can approximate g as 9.81.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + } + ], + "showMagnitude": true + }, + + { + "description": "Tension", + "content": "The force of tension acts in the positive y direction: π/2 rad. We know that the acceleration in a simple pulley system is .", + "forces": [ + { + "description": "Tension", + "magnitude": 16.68, + "directionInDegrees": 90, + "component": false + } + ], + "showMagnitude": true + } + ] + }, + "suspension": { + "question": "A 1kg weight is attached to two rods hanging from 45° angles from the ceiling. The system is in equilibrium, i.e. the weight does not move. What are the magnitudes and directions of the forces acting on the weight?", + "steps": [ + { + "description": "Forces", + "content": "There are three force acting on the red weight: the force of gravity, the force of tension from the left rod, and the force of tension from the right rod.", + "forces": [ + { + "description": "Gravity", + "magnitude": 9.81, + "directionInDegrees": 270, + "component": false + }, + { + "description": "Left Tension", + "magnitude": 6.94, + "directionInDegrees": 135, + "component": false + }, + { + "description": "Right Tension", + "magnitude": 6.94, + "directionInDegrees": 45, + "component": false + } + ], + "showMagnitude": false + } + ] + } +} |