Julie has been hired to help paint the trim of a building, but she is not convinced of the safety of the apparatus. A 4.9 m plank is suspended horizontally from the top of the building by ropes attached at each end. She knows from previous experience that the ropes being used will break if the tension exceeds 1.0 kN. Her 75 kg boss dismisses Julie's worries and begins painting while standing 1.0 m from the end of the plank. If Julie's mass is 59 kg and the plank has a mass of 21 kg, then over what range of positions can Julie stand to join her boss without causing the ropes to break?
0 < x <
To determine the range of positions where Julie can stand without causing the ropes to break, we need to analyze the forces acting on the plank.
First, let's consider the forces acting at each end of the plank. At the left end, we have the tension force exerted by the left rope, and at the right end, we have the tension force exerted by the right rope. There is also the weight of the plank acting downward at its center.
Since the plank is in equilibrium (not accelerating), the sum of the forces acting on it must be zero.
Let's start by calculating the weight of Julie, her boss, and the plank:
Weight of Julie = mass of Julie × acceleration due to gravity
Weight of Julie = 59 kg × 9.8 m/s^2 = 578.2 N
Weight of boss = mass of boss × acceleration due to gravity
Weight of boss = 75 kg × 9.8 m/s^2 = 735 N
Weight of plank = mass of plank × acceleration due to gravity
Weight of plank = 21 kg × 9.8 m/s^2 = 205.8 N
Now, let's consider the forces acting at each end of the plank:
At the left end (which is x = 0), the tension force exerts an upward force.
At the right end (which is x = length of the plank), the tension force exerts a downward force.
Since the plank is symmetric, the weight of the plank acts at its center, which is at half its length.
Now, let's analyze the forces acting on the plank when Julie is standing at a distance x from the left end:
1. The tension force on the left side is pulling up with a force of T1.
2. The tension force on the right side is pulling down with a force of T2.
3. The weight of the plank is acting downward through its center with a force of W_plank.
4. The weight of Julie is acting downward at a distance (x - 1.0 m) from the left end with a force of W_Julie.
5. The weight of the boss is acting downward at a distance (length of the plank - x) from the left end with a force of W_boss.
To ensure the ropes do not break, the maximum tension force we can have is 1.0 kN (or 1000 N).
The sum of forces in the vertical direction (Y-axis) should be zero:
ΣF_y = T1 + T2 + W_plank + W_Julie + W_boss = 0
T1 + T2 + 205.8 N + 578.2 N + 735 N = 0
Now, let's find the tension forces T1 and T2 in terms of x:
Since the plank is in equilibrium, the tension forces at each end of the rope are equal:
T1 = T2
Now, substitute T2 = T1 into the equation:
2T1 + 205.8 N + 578.2 N + 735 N = 0
2T1 = -205.8 N - 578.2 N - 735 N
2T1 = -1519 N
T1 = -759.5 N
Since we know that the tension force cannot be negative, this means that the position x where Julie can stand should be such that T1 (or T2) does not exceed 1000 N.
Thus, the range of positions where Julie can stand without causing the ropes to break is:
0 < x < (maximum value of x)
To find the maximum value of x, we need to determine the position where the tension force equals 1000 N:
2T1 + 205.8 N + 578.2 N + 735 N = 0
2T1 + 1519 N = 0
2T1 = -1519 N
T1 = -759.5 N
Since the maximum tension force is 1000 N, we set T1 (or T2) equal to 1000 N and solve for x:
2T1 + 205.8 N + 578.2 N + 735 N = 0
2(1000 N) + 205.8 N + 578.2 N + 735 N = 0
2000 N + 519 N = -735 N - 205.8 N
2519 N = -940.8 N
Now, let's solve for x by dividing both sides of the equation by -1519 N:
x = -940.8 N / -1519 N
x ≈ 0.619 m
Therefore, the range of positions where Julie can stand without causing the ropes to break is:
0 < x < 0.619 m