A boy and a girl carry a 12.0 kg bucket of water by holding the ends of a rope with the bucket attached at the middle. FI tehre is an angle of 100.0 degrees between the two segments of the rope, what is the tension in each part?
I do not understand the work
Fw = mg = (12.0 kg)(9.80 s^-2 m0
= 118 N
2^-1 Fw = 2^-1 118 N = 59.0 N
cos 50.0 = T^-1 59.0 N (T = the tension in the rope)
T = (cos 50.0)^-1 59.0 N = (.643)^-1 59.0 N = 91.8 N
why did they cut the angle and half and the the 118 N
Thanks!
Fw represents the weight equal to mass times the acceleration due to gravity,
mg=12*9.8=117.6 N
In fact, the triangle of forces consists of an isosceles triangle with two equal side of T, and the base equals to 118 N.
The solution of the triangle is simplified by dividing it into two right triangles, the base thus equals half of 118=59 N.
The angle of 100.0 degrees between the two segments of the rope allows us to split the weight of the bucket into two components, one acting on each segment of the rope. By cutting the angle in half, we can work with a right triangle and use trigonometric functions to find the tension in each part.
The weight of the bucket, Fw, is equal to the gravitational force acting on it, which can be calculated using the equation Fw = mg, where m is the mass of the bucket (12.0 kg) and g is the acceleration due to gravity (9.8 m/s^2). Thus, Fw = (12.0 kg)(9.8 m/s^2) = 118 N.
Next, we need to find the horizontal component of the weight acting on each part of the rope. Since the angle between the two segments is 100.0 degrees, the angle formed with the vertical is 50.0 degrees. We can use trigonometric functions to calculate this component.
The horizontal component can be found using the equation T*cos(50.0) = Fw/2, where T is the tension in the rope and Fw/2 is half of the weight of the bucket. Rearranging the equation, we have T = (Fw/2) / cos(50.0).
Substituting the values, we get T = (118 N / 2) / cos(50.0) = 59.0 N / cos(50.0).
The final step involves evaluating cos(50.0) to get its numerical value, which is approximately 0.643. Plugging this back into the equation, we have T = (0.643^-1) * 59.0 N = 91.8 N.
Therefore, the tension in each part of the rope is approximately 91.8 N.
To understand how the tension in each part of the rope is calculated, let's break down the problem step by step:
1. First, we calculate the weight of the bucket of water using the formula Fw = mg, where Fw is the weight, m is the mass, and g is the acceleration due to gravity. In this case, the mass of the bucket is given as 12.0 kg, and the value of acceleration due to gravity, g, is approximately 9.80 m/s².
Fw = (12.0 kg)(9.80 m/s²)
= 117.6 N (rounded to 118 N)
2. In order to find the tension in each part of the rope, we divide the weight in half because the force is being exerted in two different directions, one on each segment of the rope.
Fw/2 = 118 N / 2
= 59.0 N
3. Now, let's consider the angle between the two segments of the rope, which is given as 100.0 degrees. We can calculate the tension using trigonometry.
Since we have the adjacent side (the tension) and the hypotenuse (59.0 N), we can use the cosine function to find the tension. The formula is cos θ = adjacent/hypotenuse.
Tension (T) = (cos 50.0)⁻¹ * 59.0 N
= (0.643)⁻¹ * 59.0 N
= 91.8 N
Therefore, the tension in each part of the rope is approximately 91.8 N.