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Mechanics

Hi all,

Right, I cant seem to get this one correct, the question is:
Two forces, each of magnitude 5N, act at the point P, as shown in the diagram. Their resultant also has magnitude 5N. By drawing a vector triangle to represent the two forces and their resultant, find the angle between the two forces.
heres a rough description of the diagram, one 5N force is on the bearing 090, along the x axis basically, and the other is (after looking the answer i now know this) on the bearing 330. i need to find the angle between these two forces. (the smaller angle)
using a^2=b^2+c^2-2bcCosA you get cosA=1/2 which gets you 60 degrees. but the answer is 120 degrees, so i somehow need to get cosA=-1/2... but how? does the one vector on the bearing 330, so essentially in the other direction, make it a -5 quanitity when put into the formula for the angle or?
i know that seems very confusing, any help would be great!
Original post by TMCkins
Hi all,

Right, I cant seem to get this one correct, the question is:
Two forces, each of magnitude 5N, act at the point P, as shown in the diagram. Their resultant also has magnitude 5N. By drawing a vector triangle to represent the two forces and their resultant, find the angle between the two forces.
heres a rough description of the diagram, one 5N force is on the bearing 090, along the x axis basically, and the other is (after looking the answer i now know this) on the bearing 330. i need to find the angle between these two forces. (the smaller angle)
using a^2=b^2+c^2-2bcCosA you get cosA=1/2 which gets you 60 degrees. but the answer is 120 degrees, so i somehow need to get cosA=-1/2... but how? does the one vector on the bearing 330, so essentially in the other direction, make it a -5 quanitity when put into the formula for the angle or?
i know that seems very confusing, any help would be great!


If the two forces and the resultant all have the same magnitude then the triangle of forces must be equilateral.
In the original diagram look at your two vectors: They are both going away from the same point. At this stage make a guess - the angle between them is certainly more than 90 and less than 180 degrees.
When drawing a vector triangle and the resultant, remember that adding vectors is 'tip to tail' (imagine the two vectrs are the scenic route and the resultant is the short cut)
Now, your vector triangle has sides of magnitude 5N. Use a bit of geometry and compare with your original diagram.
PS 120 degrees is correct!
Reply 4
Original post by Mr T Pities You
PS 120 degrees is correct!


i know from the diagram that 120 degree is obvs the right answer, i just cant get their mathematically?
Reply 5
CosineA=n, (180-n),..
You have to chose which one is appropriate


Posted from TSR Mobile
I hope this makes my comments clearer!
Reply 7
Original post by tmorrall
CosineA=n, (180-n),..
You have to chose which one is appropriate


Posted from TSR Mobile


Cheers :wink:
Reply 8
Original post by Mr T Pities You
I hope this makes my comments clearer!


ahh, thanks! thats great :biggrin:
Original post by Mr T Pities You
I hope this makes my comments clearer!


up to a point... it is upside down :beard:
Original post by the bear
up to a point... it is upside down :beard:

Hey, I'm not doing all the hard work :wink:

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