Finding resultant acceleration of a sailboat and tension in a wire

[bigslowy]

These problems are really giving me fits, I would really appreciate some help.

A 12200 kg sailboat experiences an eastward
force of 27400 N due to tide pushing its hull
while the wind pushes the sails with a force of
56200 N directed toward the northwest (45±
westward of North or 45± northward ofWest).
What is the magnitude of the resultant ac-
celeration of the sailboat?



I'm using Fr= square root of (Frx squared)+(Fry squared) but I'm still not getting it right. I'm thinking that maybe I drew the free body diagram wrong. Any help?

Now for this one

The distance between two telephone poles is
32 m. When a 1:15 kg bird lands on the
telephone wire midway between the poles, the
wire sags 0:182 m.
The acceleration of gravity is 9:8 m=s2 :
How much tension in the wire does the
bird produce? Ignore the weight of the wire.

So far I've drawn the diagram and got a hypotenuse of 16.001 m, but I'm unsure how to finish the problem.

Any help would be much appreciated.

 

[Leong]

1. Find the magnitude using the cosine law.
2. use F=ma to find the magnitude of a.
It is easier to use the unit vector i and j. but i am afraid you haven't learned that yet. so we will approach it geometrically.

 

[Simonnava]

First problem (sail boat)

check the attachment

Fx = F(east) - F(northwest)Xcos45 = -12339 N
Fy = F(northwest)Xsin45 = 39739 N
Fnet = Squareroot of (Fx^2 + Fy^2)
then, Fnet = ma
a = Fnet/mass = 3.4 m/s

 

[Simonnava]

2nd problem

ok the hyp that you got is good.
a^2 + b^2 = c^2
a= 16
b = .182
c = 16.001 using the formula. Good you got until this point.

2nd step= get the angle theta. (see diagram)

90-tan (16/.182) = theta


step 3 = Get the tension by adding the Forces on the y axes. See diagram

 

[Leong]

Free body diagram : Attached file.

 

[Simonnava]

of course, I didn't draw the other tension on the other side because is understood.