Exploration 1: Balance a Mobile
The center of gravity is at the same location as the center of mass for systems in which the acceleration due to gravity is virtually equal for all points in the system. Therefore, in this Exploration we will talk about and calculate the center of mass. Restart.
The goal when making a mobile is for the center of mass to be below the string attached to the ceiling. Otherwise, there will be a net torque on the mobile about its center of mass until this condition is met. Consider a mobile made of two blocks as shown in the animation. The mass of the blue block is \(0.050\text{ kg}\). Assume that the rods and string connecting the blocks are light enough to be neglected (position is given in meters).
- What is the mass of the green block in order for the mobile to be balanced in Animation 1?
To answer this question, consider the conditions of equilibrium. The net torque on the horizontal rod about the point where it is attached to the string connected to the ceiling must equal zero. Therefore, the magnitude of the torque on the rod due to the tension in the right string must be equal to the magnitude of the torque on the rod due to the tension in the left string.
- Suppose you want to replace the green block with another two-block system just like the first one, but with less massive blocks as shown in Animation 2. What are the masses of the red and orange blocks?
This is very similar to the previous question, except that you are applying the conditions of equilibrium to the rod connecting the red and orange blocks. However, when you solve the equation, net torque \(= 0\), you have a problem. The masses of BOTH the red and orange blocks are unknown. Before, for the green and blue blocks, you knew the mass of the blue block and could solve for the mass of the green.
- You need another relationship between the red and orange blocks to help you out. Since we replaced the green block with the red and orange ones, how are the masses of these three blocks related?
- Ok, this is getting fun. Suppose that you now replace the orange block with another identical two-block system as shown in Animation 3. What are the masses of the yellow and purple blocks?
- Where is the center of mass of the system of four blocks?
Since you know the masses of the blocks, measure the \(x,\: y\) coordinates of each block and calculate the coordinates of the center of mass.
- Now click on the animation to locate the point that you just calculated. Is it directly beneath the string that connects the mobile to the ceiling? It should be!
Note
Note that adding a new system of blocks to the left end each time did not change the \(x\) coordinate of the center of mass; however, it did cause the \(y\) coordinate to decrease since each system of blocks hung lower and lower. However, shifting the \(y\) coordinate of the center of mass did not change the equilibrium status of the mobile.
Exploration authored by Aaron Titus.
Exploration 2: Static Friction on a Horizontal Beam
You hold a piece of wood by pushing it horizontally against a wall as shown in the animation (position is given in meters). Restart.
- What forces act on the wood? Draw a free-body diagram for the wood showing the forces at their proper locations. Compare your diagram to the one shown in Animation 2.
- What is the force in the \(+y\) direction that counteracts the weight of the wood in this example? Note that this force is parallel to the surface of the wall and wood where they are in contact.
- In this case, do you know if the static frictional force of the wall on the wood is equal to its maximum value?
In Animation 3 you can adjust the magnitude of the push by clicking and dragging on the white circle at the tip of the vector representing the force of your hand on the wood. The maximum frictional force (shown as a \(\color{red}\text{red vector}\)) adjusts accordingly. At the instant where the actual frictional force (\(\text{black vector}\)) equals the maximum frictional force (\(\color{red}\text{red vector}\)), the beam will still be in equilibrium. In this case this is the least force that you can push the wood and have it remain in equilibrium. If you push it with less force, the meter stick will fall.
In Animation 4 you can adjust the magnitude of the push so that it is less than the minimum push required for the wood to remain in equilibrium. If the maximum possible static frictional force is less than the actual frictional force needed for equilibrium, the piece of wood will fall. In the animation, if you adjust \(f_{\text{s max}} < f_{s}\), the resulting situation depicted in the animation is unphysical since the piece of wood will actually fall.
Exploration authored by Aaron Titus.
Exploration 3: Distributed Load
A box sits on a board of negligible weight. Two supports exert forces on the left and right ends of the board (position is given in meters). The arrows represent the relative sizes of the two force vectors, but their length does not represent their actual magnitudes. Restart.
- How does the force of each support on the board depend on where the box is located? You can drag the box from left to right to view the forces of the supports on the board.
Suppose the box is exactly halfway between the center and the right support.
- What is the ratio of the magnitude of the force of the right support on the board to the magnitude of the force of the left support on the board?
Consider a situation in which the board has significant weight.
- What then might the force vectors look like when the box is sitting above one of the supports, say for example the left support? Check using Animation 2.
- What is the ratio of the weight of the board to the weight of the box in this case?
Exploration authored by Aaron Titus.
Exploration 4: The Stacking of Bricks
How can you stack four uniform bricks (or blocks or meter sticks) one on top of another so that they extend as far over a table as possible and yet remain stable? Rules: You can drag and move bricks horizontally with your mouse (position is given in tenths of inches; hence each brick is a foot long). The stability of each brick is color coded:
- \(\color{green}\text{green:}\) the brick is in stable equilibrium
- \(\color{yellow}\text{yellow:}\) the center of gravity of the brick, or a group of bricks, is right above the edge of the supporting brick
- \(\color{red}\text{red:}\) the brick is unstable; it will fall in real life.
The center of gravity for each brick is shown as a small blue dot. The current mouse position (relative to the top left edge of the table) is shown in the upper part of the animation in the Text Field. If you press the "show c.g." button, the center of gravity for the brick subsystems (top brick, top two bricks, top three bricks and all four bricks, respectively) will be shown as a small circle with an arrow. The length of the arrow is proportional to the gravitational force for each balanced subsystem. In addition,
- \(\color{red}\text{The left edge of each brick is in red}\)
- \(\color{blue}\text{The position of the center of gravity of each brick is in blue}\)
- \(\text{The position of the center of gravity is in black when "show c.g." is selected}\)
- What is the stability condition for an object overhanging a table?
- How should each brick be positioned relative to the brick underneath it? Be explicit and refer to the center of gravity. Hint: Start at the top and work down.
- Can the top brick have its entire length beyond the edge of the table?
- Challenge: Come up with a mathematical description for the overhang of each individual brick and the total overhang.
Exploration authored by Fu-Kwun Hwang and Mario Belloni.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.
Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.
