This is the problem:


"The figure below shows a 25-foot sailboat. The mast is a uniform 119-kg pole that is supported on the deck and held fore and aft by wires as shown. The tension in the forestay (wire leading to the bow) is 840 N. Determine the tension in the backstay (wire leading aft) and the normal force that the deck exerts on the mast. (Assume that the frictional force the deck exerts on the mast to be negligible.)"

As you can see, I got it wrong the first few times. This is what I tried

1) I first determined the force of the pole as 1167.4 N
2) I determined using trig that the angle of the front rope from x plane is 240.7 degrees
3) Now I have:

Front rope is 840 N @ 240.7 degrees
The mast is 1167 N @ 270 degrees
The back rope is (?)N @ 315 degrees

4) I solved for vector components and came up with two equations to solve for the x and y components of the tension in the back rope:

 -411 N + (x)cos(315) N = 0
 -732.5 N + (y)sin(315) N + 1167.4 N = 0

 x component of back rope is 581 N
 y component of back rope is 615 N

I don't know what to do from here, I added the two components but that value of 1196 was wrong according to the application.


closed as off-topic by David Z Apr 7 '16 at 23:40

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  • $\begingroup$ Comments to the question (v1): (a) Your screen capture is unreadable -- perhaps crop to only the picture and include the question in your text. (b) I don't see where you are asking a conceptual question --- you seem to have "done some equations" and gotten stuck. $\endgroup$ – rob Apr 7 '16 at 23:17
  • $\begingroup$ Fixed. Sort of, I'm new to this. $\endgroup$ – V Ling Apr 7 '16 at 23:32
  • $\begingroup$ Hint: You've correctly found the $x$-component of the force from the forestay. What is the total force in the $x$-direction? $\endgroup$ – rob Apr 7 '16 at 23:46
  • $\begingroup$ se.u.94y.info may help $\endgroup$ – barrycarter Apr 8 '16 at 13:36

The vertical force is the sum of the weight of the mast (119*9.81) plus the sum of the vertical components of tension.

The back tension is found by equating the horizontal components (so the mast has no net horizontal force on it).

Draw all the vectors in a force diagram and follow the above - and you will get there.


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