Tension in string at a point on a frame

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LeafMuncher
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Homework Statement


Diagram of a frame with a pulley at point A, a rope running over the pulley to a 65 kg block hanging at the right, and labelled Figure 3 below the structure.

Hi all. What concerns me here is that it's worth 10 marks, but the solution I've tried only takes 2 steps. Am I missing some relation between the frame support and the pulley altering the components of the tension, or is the solution really just a basic trig conversion?

Homework Equations


F = mg
Tx = mg*sin(theta)

The Attempt at a Solution


using the above conversions I get the tension as 65kg*9,8m/s^2 = 637N
Then just convert using the 1/1 ratio as 45deg, giving Tx and Ty = 637*sin(45) = 450.4N
 
LeafMuncher said:

Homework Statement


Line diagram labelled Figure 3 of a frame with points A and B, a pulley at B with a rope running over it to a 65 kg block hanging on the right, and supports at the base.

Hi all. What concerns me here is that it's worth 10 marks, but the solution I've tried only takes 2 steps. Am I missing some relation between the frame support and the pulley altering the components of the tension, or is the solution really just a basic trig conversion?

Homework Equations


F = mg
Tx = mg*sin(theta)

The Attempt at a Solution


using the above conversions I get the tension as 65kg*9,8m/s^2 = 637N
Then just convert using the 1/1 ratio as 45deg, giving Tx and Ty = 637*sin(45) = 450.4N
Your solution is correct. Good job.