How Do You Calculate the Tension in a Rope Holding a Suspended Box at an Angle?

[fishingspree2]

Homework Statement

A 2kg box is suspended to the ceiling using only one rope. A horizontal force keeps the rope at 37 degrees from the vertical. Find the tension of the rope

Homework Equations

w = mg
F = ma

The Attempt at a Solution

my understanding:
http://img507.imageshack.us/img507/8840/diagdd2.jpg
sum of y = Tsin37 - 2*9.8 = 0
T = 32.5 N
the correct answer is 14.7
please help

 

[PhanthomJay]

37 degrees from the vertical...you show horizontal...

 

[baba89]

Your attempt at a solution is almost correct. However, there are a few errors in your calculations.

Firstly, you have used the wrong angle in the trigonometric equation. The angle between the rope and the vertical should be 90 degrees minus 37 degrees, which is equal to 53 degrees. So the correct equation should be Tsin53 - 2*9.8 = 0.

Secondly, you have used the wrong value for the weight of the box. The weight of the box is equal to its mass multiplied by the acceleration due to gravity, which is 9.8 m/s^2. So the correct weight of the box is 2*9.8 = 19.6 N.

Putting these values into the equation, we get Tsin53 - 19.6 = 0.

Solving for T, we get T = 19.6/sin53 = 25.3 N.

Therefore, the tension in the rope is 25.3 N. This is the correct answer of 14.7 N, as the question states that a horizontal force is keeping the rope at an angle, so the tension in the rope will be greater than the weight of the box. It is important to carefully read the question and use the correct values in your calculations.