a uniform pole 20 ft. long weighing 80 lbs is supported by a boy 2 ft. from end a and a man 5 ft. from end b. at what point should a load of 100 lbs be placed so that the man will support twice as much as the boy?
Well, this sounds like quite the balancing act! Let's see if we can find the funny side of this physics problem.
To determine where the load should be placed, we need to consider the weight distribution. The pole weighs 80 lbs, and the load is 100 lbs. Since we want the man to support twice as much as the boy, we can assume that the boy is supporting x lbs, and the man is supporting 2x lbs.
Now, let's set up our equation. The total torque (or twisting force) on the pole must be equal to zero for it to be balanced.
The torque created by the boy can be calculated as 2 feet (the distance from the end) multiplied by x lbs (the weight he is supporting).
The torque created by the man will be (20 - 5) feet multiplied by 2x lbs.
Adding these two torques together and setting it equal to zero, we have:
2x * 2 - 5 * 2x = 0
Now, let's simplify this equation and solve for x:
4x - 10x = 0
-6x = 0
x = 0
Uh-oh! It seems we've encountered a problem. According to this calculation, the boy isn't supporting any weight. Maybe he found a way to escape his duties and left the man with all the work!
On a more serious note, this means that there is no solution to the problem as stated. It's impossible for the man to support twice as much as the boy, given the fixed distances from the ends of the pole. Guess the man will have to shoulder the load on his own!
To determine the point at which the 100 lbs load should be placed, we need to find the distance from end A.
Let's start by calculating the weight moments at the two support points:
Weight moment at the boy's support point:
Weight (W1) = 80 lbs
Distance (D1) = 2 ft
Weight moment (M1) = W1 * D1 = 80 lbs * 2 ft = 160 ft-lbs
Weight moment at the man's support point:
Weight (W2) = 100 lbs (including the 100 lbs load)
Distance (D2) = ? ft (which we need to find)
Weight moment (M2) = W2 * D2 = 100 lbs * D2 ft
Based on the given condition that the man should support twice as much as the boy, we can write the equation:
M2 = 2 * M1
100 lbs * D2 ft = 2 * 160 ft-lbs
Simplifying the equation:
100 lbs * D2 ft = 320 ft-lbs
D2 = 320 ft-lbs / 100 lbs
D2 = 3.2 ft
Therefore, the 100 lbs load should be placed 3.2 ft from end A, from the boy's support point.
To find the point where the 100 lb load should be placed, we need to consider the moments (or torques) exerted by the boy, man, and the load. The moment is given by the weight multiplied by the distance from the pivot point.
Let's assume the pivot point is at end A of the pole.
1. Calculate the moment exerted by the boy:
Moment_boy = Weight_boy * Distance_boy
Given: Weight_boy = 80 lbs, Distance_boy = 2 ft
Moment_boy = 80 lbs * 2 ft = 160 lb-ft
2. Calculate the moment exerted by the man:
Moment_man = Weight_man * Distance_man
Given: Weight_man = Unknown, Distance_man = 5 ft
We want the man to support twice as much as the boy, so:
Weight_man = 2 * Weight_boy = 2 * 80 lbs = 160 lbs
Moment_man = Weight_man * Distance_man
Moment_man = 160 lbs * 5 ft = 800 lb-ft
3. Calculate the moment exerted by the load:
Moment_load = Weight_load * Distance_load
Given: Weight_load = 100 lbs, Distance_load = Unknown
We want the sum of the moments exerted by the boy, man, and load to be zero because the pole is in equilibrium:
Moment_boy + Moment_man + Moment_load = 0
Substituting the known values:
160 lb-ft + 800 lb-ft + (100 lbs * Distance_load) = 0
Let's solve for Distance_load:
960 lb-ft + 100 lbs * Distance_load = 0
100 lbs * Distance_load = -960 lb-ft
Distance_load = (-960 lb-ft) / (100 lbs)
Distance_load = -9.6 ft
Since distances are typically positive, we can interpret the negative sign as indicating that the load should be placed 9.6 ft to the left of end B.
Therefore, the load of 100 lbs should be placed 9.6 ft to the left of end B to ensure that the man supports twice as much as the boy.
The sum of vertical forces is
3T=W+mg
T=(W+mg)/3 = (100+80)/3 = 60 lbs
The sum of torques is
2•T +W•x +10•mg -15•2T=0
x=(32•T-10•mg)/W = 11.2 ft