A physics student of mass mm = 100 kg gets a summer job painting houses. On his first project, he builds a platform using large pine board with a mass of mb = 60 kg The board has an overall length of L = 8.5 meters and set on two 1 meter tall saw horses so that l = 2.25 meters overhangs each side as shown above.
How far from the left end of the beam can the painter stand before the board (and painter) begin to tip over?
x = m
HELP: In order to unbalance the beam, the beam would lift up from the support at point A. Therefore, you should pivot about point B. Also, notice that if the board lifts up at point A, that the normal force at point A is zero.
HELP: Using the point B as the pivot, there are two forces that contribute to the torque, the force of the man's weight, and the force of the weight of the beam.
Later that day, after thinking about how cool rotational dynamics really is, the student decides to conduct an experiment. He removes one of the supports and places the other one 1/3 of the way from the left edge. Standing at the end of the board, he has his girl friend place paint cans, each of mass mc = 1.94 kg, on the opposite end. How many cans will the girl have to place on the board to provide the best balance? (You may neglect the small length of the board that both the man and the cans occupy. Assume both are points at the ends of the board.)
Number of cans = cans
HELP: For the board to remain stable, the torques all have to add up to zero. We could pick any point we wanted to to sum the torque, but some points might make the math easier. What would be a good point to sum the torques about?
HELP: How many forces are there causing torques?
With the HELP they have already provided, they have explained how to do the problem. Please show your work for further assistance.
can you write me some kind of a equation or explain to me in a way which i can understand. because what they wrote is just flying over my head and i have already spent a good amount of time trying to figure these out. so please help
nvm i got it, thanks for the help guys.
To solve the first part of the problem, we need to find the maximum distance from the left end of the board the painter can stand before it tips over.
Let's start by calculating the torques acting on the board when it's on the verge of tipping. We'll take point B as the pivot.
The torque due to the painter's weight is given by the equation:
Torque_painter = weight_painter × distance_painter
The painter's weight is the gravitational force, which can be calculated as:
Weight_painter = mass_painter × gravity
The torque due to the board's weight is given by the equation:
Torque_board = weight_board × distance_board
The board's weight is the gravitational force, which can be calculated as:
Weight_board = mass_board × gravity
The total torque at point B is the sum of the torques from the painter and the board:
Total_torque = Torque_painter + Torque_board
Now, let's consider the forces acting at point A. When the board begins to tip, the normal force at point A becomes zero. This means that the sum of the torques at point B must also be zero.
Based on the information provided, we know the following values:
mass_painter (m) = 100 kg
mass_board (mb) = 60 kg
length_board (L) = 8.5 m
overhang (l) = 2.25 m
distance_painter (x) = ?
Using the equation for total torque, we can calculate the distance from the left end of the board that the painter can stand before it tips over:
Total_torque = Torque_painter + Torque_board
0 = Weight_painter × distance_painter + Weight_board × distance_board
0 = (mass_painter × gravity) × x + (mass_board × gravity) × (L - x)
Now, we can substitute the known values and solve for x:
0 = (100 kg × 9.8 m/s²) × x + (60 kg × 9.8 m/s²) × (8.5 m - x)
Simplifying the equation:
0 = 980x + 588 - 588x
Combine like terms:
-392x = -588
Divide both sides by -392:
x = 1.5 meters
Therefore, the painter can stand a maximum distance of 1.5 meters from the left end of the board before it begins to tip over.
Now, let's move on to the second part of the problem.
In this part, we need to determine the number of paint cans (mc) that the girl needs to place on the opposite end of the board to provide the best balance.
To achieve balance, the torques acting on the board must add up to zero. Let's choose the pivot point at the end of the board, opposite the girl's placement.
The torque due to the painter's weight is equal to:
Torque_painter = weight_painter × distance_painter
The torque due to each paint can's weight is equal to:
Torque_can = weight_can × distance_can
The total torque at the pivot point is the sum of all the individual torques:
Total_torque = Torque_painter + (Torque_can × Number_of_cans)
Since the board is in equilibrium, the total torque must be zero. Therefore, we set up the equation:
0 = (mass_painter × gravity) × distance_painter + (mass_can × gravity) × distance_can × Number_of_cans
Using the given values:
mass_painter (m) = 100 kg
mass_can (mc) = 1.94 kg
distance_painter (x) = (2/3) × length_board (L)
Substituting the values and solving for the number of paint cans:
0 = (100 kg × 9.8 m/s²) × (2/3) × 8.5 m + (1.94 kg × 9.8 m/s²) × 1.5 m × Number_of_cans
Simplifying the equation:
0 = 647 - 28.57 Number_of_cans
Rearranging the equation:
28.57 Number_of_cans = 647
Divide both sides by 28.57:
Number_of_cans = 22.63
Since we can't have a fraction of a paint can, we need to round up to the nearest whole number. Therefore, the girl needs to place 23 paint cans on the board to provide the best balance.
Therefore, the answer to the problem is:
The painter can stand a maximum distance of 1.5 meters from the left end of the board before it begins to tip over.
The girl needs to place 23 paint cans on the board to provide the best balance.