Calculate the tension in the two ropes if the person is motionless. Person weighs 76kg and 15 degree angle rope one and 10 degree angle rope 2
To calculate the tension in the two ropes, we need to use the basic principles of physics, specifically the forces acting on the person.
Let's denote the tension in rope one as T1 and the tension in rope two as T2.
First, we need to resolve the weight of the person into its components along the ropes. The vertical component (perpendicular to the ropes) will be counteracted by the tension in the ropes, while the horizontal component won't have any effect on the tension.
1. Calculate the vertical component of the weight:
Vertical Component = Weight * cos(angle)
Vertical Component = 76 kg * 9.8 m/s^2 * cos(15°)
2. Calculate the horizontal component of the weight:
Horizontal Component = Weight * sin(angle)
Horizontal Component = 76 kg * 9.8 m/s^2 * sin(15°)
Now, we can write equations for the vertical and horizontal forces:
Equation 1: T1 + T2 = Vertical Component
Equation 2: Horizontal Component = 0
Substituting the values:
Equation 1: T1 + T2 = 76 kg * 9.8 m/s^2 * cos(15°)
Equation 2: Horizontal Component = 76 kg * 9.8 m/s^2 * sin(15°)
Since the person is motionless, the horizontal component is zero. Therefore, equation 2 becomes:
0 = 76 kg * 9.8 m/s^2 * sin(15°)
Solving equation 2 for T2:
T2 = -T1
Substituting T2 in equation 1:
T1 + (-T1) = 76 kg * 9.8 m/s^2 * cos(15°)
Simplifying equation 1:
0 = 76 kg * 9.8 m/s^2 * cos(15°)
Thus, the tension in rope one (T1) and rope two (T2) will be zero if the person is motionless.
To calculate the tension in the two ropes, we need to analyze the forces acting on the person. We can break down the weight of the person into vertical and horizontal components.
1. Vertical Component:
The vertical component of the weight can be calculated using the formula: weight * cos(angle).
Vertical component = 76 kg * 9.8 m/s^2 * cos(15 degrees) = 731.518 N
2. Horizontal Component:
The horizontal component of the weight can be calculated using the formula: weight * sin(angle).
Horizontal component rope one = 76 kg * 9.8 m/s^2 * sin(15 degrees) = 192.802 N
Horizontal component rope two = 76 kg * 9.8 m/s^2 * sin(10 degrees) = 129.789 N
Now, let's analyze the forces acting in each rope.
For rope one:
Let T1 be the tension in the rope one. Since the person is motionless, the sum of the vertical forces must be zero.
Vertical forces: T1 * cos(15 degrees) = 731.518 N
Solving the above equation, we find:
T1 = 731.518 N / cos(15 degrees) = 766.413 N (rounded to three decimal places)
For rope two:
Let T2 be the tension in rope two. Again, since the person is motionless, the sum of the vertical forces must be zero.
Vertical forces: T2 * cos(10 degrees) = 731.518 N
Solving the above equation, we find:
T2 = 731.518 N / cos(10 degrees) = 748.383 N (rounded to three decimal places)
Therefore, the tension in rope one is approximately 766.413 N, and the tension in rope two is approximately 748.383 N when the person is motionless.
M*g = 76 * 9.8 = 745 N.=Wt. of person.
T1*Cos(180-15) + T2*Cos10 = -745*cos270.
-0.966T1 + 0.985T2 = 0.
T2 = 0.981T1
-500i + T1*sin(180-15) + T2*sin10 = 0.
Replace T2 with 0.981T1:
-500i + T1*sin165 + 0.966T1*sin10 = 0.
-500i + 0.259T1 + 0.168T1 = 0.
-500i + 0.427T1 = 0.
0.427T1 = 500i.
T1 = 1172i = 1172N.[90o].
T2 = 0.981*1172i = 1149i = 1149N[90o].
Correction:
M*g = 76*9.8 = 745 N. = Wt. of person.
T1*Cos(180-15) + T2*Cos10 = -745*Cos270.
-0.966T1 + 0.985T2 = 0.
T2 = 0.981T1.
-745 + T1*sin(180-15) + T2*sin10 = 0.
Replace T2 with 0.981T1:
-745 + T1*sin(180-15) + 0.981T1*sin10=0.
-745 + 0.259T1 + 0.170T1 = 0.
-745 + 0.429T1 = 0.
0.429T1 = 745.
T1 = 1735 N.
T2 = 0.981 * 1735 = 1702 N.