An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope. The rope will break if the tension in it exceeds 25000N, and our hero’s mass is 90kg.
(a) If the angle is 10o, find the tension in the rope.
(b) What is the smallest value the angle can have if the rope is not to break?
(a) Well, well, well, an adventurous archaeologist hanging in the balance! Let's find out what the tension in the rope is.
In this case, we can start by breaking down the forces acting on our hero. We have the tension in the rope pulling him upward, which we'll call T. We also have the force of gravity pulling him downward, which is given by his mass multiplied by the acceleration due to gravity (9.8 m/s^2).
Now, let's tackle some trigonometry! The tension in the rope is related to the angle theta and the weight of our hero. The vertical component of the tension T is given by T * cos(theta), and the horizontal component is given by T * sin(theta).
Since our hero is at rest and not accelerating up or down, we can set the sum of the vertical forces equal to zero. So, we have:
T * cos(theta) - (mass * gravity) = 0
Substituting the values we know, we have:
T * cos(10o) - (90 kg * 9.8 m/s^2) = 0
Now we can solve for T:
T * cos(10o) = 90 kg * 9.8 m/s^2
T = (90 kg * 9.8 m/s^2) / cos(10o)
Calculating that out, we get T ≈ 905.03N. So the tension in the rope is approximately 905.03N.
(b) Now, let's get onto the smallest value the angle theta can have, without having a broken rope show! To find this out, we need to make sure that the tension in the rope doesn't exceed 25000N.
Using the same equation as before, but this time keeping T as a variable, we have:
T * cos(theta) - (90 kg * 9.8 m/s^2) = 0
Now, we can solve for the smallest value of theta using this equation and the limit T ≤ 25000N.
theta = acos((90 kg * 9.8 m/s^2) / 25000N)
Calculating that out, we get theta ≈ 78.88o. So the smallest value the angle theta can have is approximately 78.88o.
Now, let's hope our adventurous archaeologist doesn't meet the rock bottom anytime soon! Keep that rope intact, buddy!
To solve this problem, we can use trigonometry and Newton's second law (F = ma) to find the tension in the rope and the smallest value the angle can have. Let's break it down step by step:
(a) To find the tension in the rope when the angle is 10 degrees, we can use the equation Tension = weight / cos(angle), where weight = mass * acceleration due to gravity.
Step 1: Calculate the weight of the adventurer.
Weight = mass * acceleration due to gravity
Weight = 90 kg * 9.8 m/s^2
Weight = 882 N
Step 2: Calculate the tension in the rope.
Tension = Weight / cos(angle in radians)
Angle in radians = angle in degrees * pi / 180
Angle in radians = 10 * pi / 180
Angle in radians = 0.1745 rad
Tension = 882 N / cos(0.1745 rad)
Tension ≈ 9061.02 N
Therefore, the tension in the rope is approximately 9061.02 N when the angle is 10 degrees.
(b) To find the smallest value the angle can have without breaking the rope, we need to calculate the maximum weight the rope can support without exceeding the tension limit of 25000 N.
Step 1: Calculate the maximum weight the rope can support.
Max weight = tension limit
Step 2: Calculate the smallest value of the weight.
Weight = mass * acceleration due to gravity
Weight = 90 kg * 9.8 m/s^2
Weight = 882 N
Step 3: Calculate the smallest value of the angle.
Tension = weight / cos(angle in radians)
25000 N = 882 N / cos(angle in radians)
cos(angle in radians) = 882 N / 25000 N
cos(angle in radians) ≈ 0.03528
angle in radians ≈ arccos(0.03528)
angle in radians ≈ 1.5301 rad
Angle in degrees = angle in radians * 180 / pi
Angle in degrees ≈ 1.5301 rad * 180 / pi
Angle in degrees ≈ 87.84 degrees
Therefore, the smallest value the angle can have without breaking the rope is approximately 87.84 degrees.
To solve these problems, you need to use the principles of forces and equilibrium. Here's how you can approach each part of the question:
(a) To find the tension in the rope when the angle is 10 degrees, you can use the concept of forces in equilibrium. In equilibrium, the sum of all forces acting on an object is equal to zero.
In this case, the forces acting on the archaeologist are his weight (mg) acting vertically downward and the tension in the rope (T) acting at an angle. We can resolve the force of tension into horizontal and vertical components:
T_horizontal = T * cos(angle)
T_vertical = T * sin(angle)
Since the archaeologist is at rest, the sum of the vertical forces must be zero. This means that the weight must be balanced by the vertical component of the tension:
mg = T_vertical
Substituting the values, we have:
mg = T * sin(angle)
Plugging in the given values, m = 90 kg and angle = 10 degrees, we can find the tension in the rope:
T = (90 kg) * (9.8 m/s^2) / sin(10 degrees)
(b) To find the smallest value the angle can have without breaking the rope, we need to determine the maximum tension the rope can tolerate. In this case, the maximum tension is given as 25000 N.
Using the same principle of forces in equilibrium, we can find the maximum value for tension when the angle is at its smallest. Since the tension is at its maximum, its vertical component contributes to the maximum value. Hence:
T_max = T_vertical
Applying the same equation as in part (a), we can write:
mg = T_max * sin(angle)
Rearranging the equation, we get:
angle = arcsin(mg / T_max)
Plugging in the values, m = 90 kg and T_max = 25000 N, we can find the smallest value for the angle:
angle = arcsin((90 kg * 9.8 m/s^2) / 25000 N)
That's how you can solve the two parts of the question. Remember to use the appropriate units (meters for length, newtons for force, and degrees for angles) in your calculations to get accurate results.
draw the diagrem.
each side holds half his weight.
sin10=W/2 / Tension
solve for tensions.
b. put in tension=2500
solve for theta