Nelson Thornes explanation?!
Does anyone know why the answer for 3.(a( in the Nelson Thornes Physics as level A, book chapter 7.2 is 139N?? The question is, An archer pulled a bowstring back until the two halves of the string were at 140 degrees to each other, the force needed to the string in this position was 95N. a) calculate the tension in each part of the bowstring in this position.
(edited 11 years ago)
Original post by Pin3apple
Does anyone know why the answer for 3.(a( in the Nelson Thornes Physics as level A, book chapter 7.2 is 139N??
It may widen your audience if you post the question. . .
Original post by Pin3apple
Does anyone know why the answer for 3.(a( in the Nelson Thornes Physics as level A, book chapter 7.2 is 139N?? The question is, An archer pulled a bowstring back until the two halves of the string were at 140 degrees to each other, the force needed to the string in this position was 95N. a) calculate the tension in each part of the bowstring in this position.
The tension T in each part of the bowstring will be the same. The force of 95N has to be balanced by the sum of the forces from the two parts of the string in the direction of the pull.
You won't have to worry about the forces perpendicular to the pull because the two parts of the string will cancel by symmetry.
Original post by Brister
The tension T in each part of the bowstring will be the same. The force of 95N has to be balanced by the sum of the forces from the two parts of the string in the direction of the pull.
You won't have to worry about the forces perpendicular to the pull because the two parts of the string will cancel by symmetry.
You won't have to worry about the forces perpendicular to the pull because the two parts of the string will cancel by symmetry.
What calculation would the book have used though, to get 139N. Sorry I did not explain myself well in the question?
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