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A level physics question

AD4E9C5C-BCCD-4952-A02C-AD44F7157A8E.jpg.jpeg


Hiya, im not sure how to do the last part of this question, when working it out I assumed that T was the hypotenuse because W is vertical and T makes 40 degree to horizontal so T=W/sin40. However, the mark scheme says T= Wsin40, I’m not sure why this is, I have attached a drawing of my free body diagram but I’m still not sure why this is the correct answer.
(edited 3 years ago)
Reply 1
I have just realised that the picture is pretty blurry so I have written out the question here:
(the climber is in equilibrium at this point)

One of the forces, which is assumed to be acting perpendicular to the rope is
already shown. Label this force, and add labelled arrows to the diagram to
represent the other two forces acting on the climber. Assume that the rope hanging
down from the climber exerts a negligible force on him.


(ii) The rope is at an angle of 40° to the horizontal. Calculate the tension in the rope.
Original post by Emily~3695
AD4E9C5C-BCCD-4952-A02C-AD44F7157A8E.jpg.jpeg


Hiya, im not sure how to do the last part of this question, when working it out I assumed that T was the hypotenuse because W is vertical and T makes 40 degree to horizontal so T=W/sin40. However, the mark scheme says T= Wsin40, I’m not sure why this is, I have attached a drawing of my free body diagram but I’m still not sure why this is the correct answer.


As the question states “One of the forces, which is assumed to be acting perpendicular to the rope is already shown.” and the force diagram shows the direction of the rope is perpendicular to one of the forces. The direction of the rope is also the direction of tension which cannot be hypotenuse side of a right-angled triangle. Have a look at the right-angled triangle.

I just rearranged the forces to make a triangle and solved for the adjacent otherwise you could do 638sin(40) instead.
Hope this helps

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