In a rescue, a 74.1 kg police officer is suspended by two cables, as shown in the figure. Cable A makes an angle θA = 32.0° with respect to the horizontal, and cable B, θB = 74.0°. Sketch a free body diagram. Calculate the tension in cable A.
Well, imagine the police officer being suspended by two cables. It's like a high-stakes puppet show. Cable A makes an angle of 32.0° with the horizontal, and cable B is a bit more rebellious with an angle of 74.0°. Quite the rebellious cables we have here!
Now, let's sketch a free body diagram, shall we? Picture the suspended police officer as a happy little stick figure floating in mid-air. Draw two cables, one coming from above at angle θA and the other from below at angle θB. Label the forces acting on the stick figure accordingly.
Now, to find the tension in cable A, we'll use our knowledge of trigonometry. Keep calm, it's not as scary as it sounds. We can break down the forces in the x and y directions.
In the x-direction, the tension in cable A is the only force acting, so it's equal to the horizontal component of the weight of our police officer. That would be Tension A = Weight * cos(θA).
In the y-direction, we have two forces: the vertical component of the weight and the tension in cable B. These forces must balance each other. That means Tension B = Weight * cos(θB).
Now since we know the total weight of the police officer (let's call it W), we can plug in the values and solve for Tension A.
Tension A = W * cos(θA)
But wait, we don't know the weight of the police officer! So, let's assume it's 74.1 kg for now until we get more specifics. (Don't worry, our officer won't mind a little weight fluctuation!)
Tension A = 74.1 kg * cos(32.0°)
Now go ahead and grab your calculator to find the value of Tension A.
To sketch a free body diagram, we need to consider the forces acting on the police officer. In this case, there are three forces to consider: the weight of the police officer, the tension in cable A, and the tension in cable B. The weight of the police officer can be represented by a downward force labeled "mg" (where m is the mass of the police officer and g is the acceleration due to gravity). The tension in cable A can be represented by a force labeled "T_A" that is directed at an angle θ_A = 32.0° with respect to the horizontal. The tension in cable B can be represented by a force labeled "T_B" that is directed at an angle θ_B = 74.0°.
To calculate the tension in cable A, we can use the fact that the net force in the vertical direction must be zero since the police officer is in equilibrium. This means that the vertical component of the tension in cable A must balance out the weight of the police officer.
Using trigonometry, we can determine the vertical component of the tension in cable A.
Vertical component of T_A = T_A * sin(θ_A)
Since the vertical component of T_A must balance out the weight of the police officer, we have:
T_A * sin(θ_A) = mg
Now, we can plug in the given values to calculate the tension in cable A.
m = 74.1 kg
g = 9.8 m/s^2
θ_A = 32.0°
T_A * sin(32.0°) = (74.1 kg) * (9.8 m/s^2)
T_A = [(74.1 kg) * (9.8 m/s^2)] / sin(32.0°)
T_A ≈ 446.82 N
Therefore, the tension in cable A is approximately 446.82 N.
To calculate the tension in cable A, we first need to analyze the forces acting on the police officer.
Firstly, draw a free body diagram. In this case, the diagram would include a point representing the police officer, two cables (cable A and cable B), and the force of gravity acting vertically downward on the police officer.
The force of gravity can be represented as the weight of the police officer, which is equal to the mass of the officer multiplied by the acceleration due to gravity (9.8 m/s^2):
Weight = mass × acceleration due to gravity
= 74.1 kg × 9.8 m/s^2
Next, for each cable, we need to resolve the forces into their horizontal and vertical components.
For cable A:
The horizontal component of the tension in cable A is equal to the horizontal component of the force of gravity since there are no other horizontal forces acting on the police officer.
The vertical component of the tension in cable A can be found using trigonometry. We know the angle (θA = 32.0°) and the weight of the police officer, so we can use the sine function to find the vertical component:
Vertical component of tension in cable A = Weight × sin(θA)
Now, we can calculate the tension in cable A. Since tension is a force, it is equal to the vector sum of its horizontal and vertical components.
Tension in cable A = √[(Horizontal component of tension in cable A)^2 + (Vertical component of tension in cable A)^2]
You can substitute the calculated values into the formula to find the tension in cable A.
T1*sin148 + T2*sin74 = 725 N.
Eq1: 0.53T1 + 0.96T2 = 725.
T1*Cos148 + T2*Cos74 = 0.
Eq2: -0.85T1 + 0.28T2 = 0.
-0.85T1 = -0.28T2.
T1 = 0.324T2.
In Eq1, Replace T1 with 0.324T2:
0.53*0.324T2 + 0.96T2 = 725.
0.172T2 + 0.96T2 = 725.
1.13T2 = 725.
T2 = 640.5 N.
T1 = 0.324*640.5 = 207.5 N. = Ta.