a man weighiing 7-kg lies in a hammock whose ropes make angles of 20 and 25 degrees with the horizontal. What is the tension in each rope?
the horizontal components are equal in magnitude or the guy would accelerate.
A cos 20 = B cos 25
.940 A = .906 B
A = .964 B
A sin 20 + B sin 25 = 7*9.8
.964 B (.342) + B (.423) = 68.6
I think you can solve for B and go back and get A
Rofl, I think you're in my calculus class. I have the same question for homework too.
I'm not sure if you're solving the problem using geometric or catersian vectors.
It's always best to draw out the problem. . .
Assuming you meant 70kg instead of 7kg. . .
I'll do this problem using geometric vectors.
You have a triangle with the following angles: 65, 45 and 70.
70 corresponds with |T1| (Tension 1)
65 corresponds with |T2| (Tension 2)
45 corresponds with 686 N (9.8 x 70)
Use sine law to solve for the tensions.
|T1|/sin70 = 686/sin45
|T1| = 911.6 N
|T2|/sin65 = 686/sin45
|T2| = 879.3 N
Therefore, the tensions are 911.6 N and 879.3 N
thx, i just have problems drawing the diagrams
To find the tension in each rope, we can use the concept of resolving forces into their component vectors.
Let's start by drawing a diagram of the situation. We have a hammock with a man lying on it, and two ropes attached to the hammock. The angles that the ropes make with the horizontal are given as 20 degrees and 25 degrees.
Now, we can resolve the weight of the man into its components along the two ropes. The vertical component of the weight is responsible for pulling the hammock downward, and the horizontal component is responsible for pulling the ropes outward.
First, let's calculate the vertical component of the weight of the man. We can use trigonometry to find this component:
Vertical component = weight of man * cos(angle)
Vertical component = 7 kg * cos(20 degrees)
Next, let's calculate the horizontal component of the weight of the man:
Horizontal component = weight of man * sin(angle)
Horizontal component = 7 kg * sin(20 degrees)
Now, the tension in each rope can be found by resolving the vertical and horizontal forces acting on the ropes.
The tension in the first rope (with angle 20 degrees) is given by:
Tension1 = vertical component / cos(angle)
Tension1 = (7 kg * cos(20 degrees)) / cos(20 degrees)
Simplifying, we get:
Tension1 = 7 kg
So, the tension in the first rope is 7 kg.
Similarly, we can find the tension in the second rope (with angle 25 degrees):
Tension2 = vertical component / cos(angle)
Tension2 = (7 kg * cos(25 degrees)) / cos(25 degrees)
Simplifying, we get:
Tension2 = 7 kg
So, the tension in the second rope is also 7 kg.
Therefore, the tension in each rope is 7 kg.