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Procedure 1, 2, 3 Mass of a clamp: 21.5 True mass of meter stick: _133.6 g...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move the meter stick in the support clamp until the meter stick is in equilibrium. Record this new equilibrium position as x," Using the total mass of the meter stick, calculate the clockwise and counterclockwise torques, and then calculate a percent difference. In this calculation, you will include the mass of the meter stick as if it were concentrated at its center of mass, x,...
at With the meter stick on the support stand, suspend a gram mass m = 100g...
at With the meter stick on the support stand, suspend a gram mass m = 100g the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2 = 200g on the other side of the axis. See Figure. X axis X X2 m m2 Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (Tow)...
Torques and Center of Mass. The Experiment: In this experiment, you balance a meter stick, to...
Torques and Center of Mass. The
Experiment:
In this experiment, you balance a meter stick, to balance the
meter stick, attach masses at positions until the system is in
equilibrium.
The meter stick acts as if all its mass was concentrated at its
center of mass. With the fulcrum at the center of mass, r (the
distance from the axis of rotation to the place where the force is
applied) is 0, so there’s no torque due to the meter...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the other side a counter mass m2 = 300g and adjust its position until the system is in static equilibrium. Using, Στ = 0 calculate the unknown mass m². Remove the unknown mass and determine its mass on the laboratory balance. See Figure. This is the accepted mass. Calculate % error. % Error lexperimentalacceptedy 100 accepted X axis X X2 m m, m=? X =...
the experimental part for this question is all answered there is only the last part where it says: Compute the net torque about 20 cm mark. Mass of the stick : 0.1383 kg Compu...
the experimental part for this question is all answered there is
only the last part where it says:
Compute the net torque about 20 cm
mark.
Mass of the stick : 0.1383 kg
Compute the net torque about the end of the stick.
please be sure to put detailed answer for both parts 1 and 2
thanks for the help!
Q1) Torques and equilibrium Part A: use the following methods to find the center of mass for a measuring stick...
A meter stick is found to balance at the 50.0 cm mark when placed on a...
A meter stick is found to balance at the 50.0 cm mark when placed on a fulcrum. When a 0.100 kg mass is attached at the 10.0 cm mark, the fulcrum must be move to the 34.0 cm mark for balance. a) What is the mass of the meter stick? b) How much torque does the 100 g mass exert on the meter stick around an axis at the new fulcrum (at 34.0 cm) c) How much torque does the...
2. A meter stick was pivoted at the 30 cm mark with its center of gravity...
2. A meter stick was pivoted at the 30 cm mark with its center of gravity at the 50 cm mark. If a mass of 25.29 g is hanging at the 70.35 cm mark, what mass must be hung from the 11.98 cm mark in order for the system to be in equilibrium? (Take the mass of the stick to be 98 g)
2. A meter stick was pivoted at the 30 cm mark with its center of gravity...
2. A meter stick was pivoted at the 30 cm mark with its center of gravity at the 50 cm mark. If a mass of 25.29 g is hanging at the 70.35 cm mark, what mass must be hung from the 11.98 cm mark in order for the system to be in equilibrium? (Take the mass of the stick to be 98 8)
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0...
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0 cm. The first mass of 35.0 kg is placed at the 23.0 cm mark of the meterstick. What is the location of a mass - 65.0 kg that balances the meter stick (Sum of torques
A uniform wooden meter stick has a mass of m = 799 g. A clamp can be attached to the measuring stick at any point Palong the stick so that the stuck can rotate freely about point P
A uniform wooden meter stick has a mass of m = 799 g. A clamp can be attached to the measuring stick at any point Palong the stick so that the stuck can rotate freely about point P, which is at a distance d from the zero-end of the stick as shown. Part (a) Calculate the moment of inertia in kg-m of the meter stick if the pivot point P is at the 50-cm mark. Part (b) Calculate the moment of inertia...
Suspend a mass m, =100g at or near the zero end of the meter stick. Move the meter stick in the support clamp until the meter stick is in equilibrium. Record this new equilibrium position as x," Using the total mass of the meter stick, calculate the clockwise and counterclockwise torques, and then calculate a percent difference. In this calculation, you will include the mass of the meter stick as if it were concentrated at its center of mass, x,...
at With the meter stick on the support stand, suspend a gram mass m = 100g the 15-cm mark on the meter stick. Then adjust the lever arm for a gram mass m2 = 200g on the other side of the axis. See Figure. X axis X X2 m m2 Record the mass and position x as read on the meter stick and then record the lever arms. Compute the torques and find a percent difference between the clockwise (Tow)...
Torques and Center of Mass. The
Experiment:
In this experiment, you balance a meter stick, to balance the
meter stick, attach masses at positions until the system is in
equilibrium.
The meter stick acts as if all its mass was concentrated at its
center of mass. With the fulcrum at the center of mass, r (the
distance from the axis of rotation to the place where the force is
applied) is 0, so there’s no torque due to the meter...
Place an unknown mass at the 10 cm mark of the meter stick. Suspend from the other side a counter mass m2 = 300g and adjust its position until the system is in static equilibrium. Using, Στ = 0 calculate the unknown mass m². Remove the unknown mass and determine its mass on the laboratory balance. See Figure. This is the accepted mass. Calculate % error. % Error lexperimentalacceptedy 100 accepted X axis X X2 m m, m=? X =...
the experimental part for this question is all answered there is
only the last part where it says:
Compute the net torque about 20 cm
mark.
Mass of the stick : 0.1383 kg
Compute the net torque about the end of the stick.
please be sure to put detailed answer for both parts 1 and 2
thanks for the help!
Q1) Torques and equilibrium Part A: use the following methods to find the center of mass for a measuring stick...
2. A meter stick was pivoted at the 30 cm mark with its center of gravity at the 50 cm mark. If a mass of 25.29 g is hanging at the 70.35 cm mark, what mass must be hung from the 11.98 cm mark in order for the system to be in equilibrium? (Take the mass of the stick to be 98 g)
2. A meter stick was pivoted at the 30 cm mark with its center of gravity at the 50 cm mark. If a mass of 25.29 g is hanging at the 70.35 cm mark, what mass must be hung from the 11.98 cm mark in order for the system to be in equilibrium? (Take the mass of the stick to be 98 8)
0/30 Torque and Conditions for Equilibrium. The pivot point of a meter stick is at 50.0 cm. The first mass of 35.0 kg is placed at the 23.0 cm mark of the meterstick. What is the location of a mass - 65.0 kg that balances the meter stick (Sum of torques
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