vector/forces problem?
Hey i dont know how to approach this question at first i thought it would be easy but then i realized the vectors dont add up to 0 so i cant use a triange...help would be appreciated thanks
Hey i dont know how to approach this question at first i thought it would be easy but then i realized the vectors dont add up to 0 so i cant use a triange...help would be appreciated thanks
Comments
1st vector 25N towards Horizontal (right side)
2nd vector component in horizontal direction is
- 30 cos 60=-15N
Horizontal component of 3rd vector is -10 cos 60= -5 N
Net force in horizontal direction = 25-(15+5)= 5N
Net vertical component=30 sin 60-10 sin60
=20X root 3/2
Resultant= sqr root of (5sqr+
(20X root 3/2)=sqr.root325
direction= sin inverse of 2root3
Vectors ad graphically as if you put the tail of 2nd vector at the point of 1st vector, and the tail of the 3rd vector at the point of the second. The resultant vector has its tail at the tail of the first vector and its point at the point of the 3rd vector.
that is how you visualize it.
To calculate it, you add the vectors together, using their vertical and horizontal components
To demonstrate:
!st -- horizontal component -> 25,
vertical - 0
2nd - horizontal -> 30cos(120) = -cos(60) = -15
vertical -> 30sin(120) = 30sin(60) = (15)3^(1/2)
3rd - horizontal -> 40cos(240) = -40cos(60) = -20
vertical -> 40sin240 = -40sin60 = -(20)3^(1/2)
The resulting vector has the following components.
horizontal -> +25 + (-15) + (-20) = -10
vertical -> 0 + (15)3^(1/2) + (-20)3^(1/2) = -(5)3^(1/2)
I will leave it to you to find the angle of this resultant vector from its component.
Also, keeping track of + and - signs were always a weakness of mine, so check carefully.
resolve the forces in the horizontal and verticle directions
assume that we are talking about a point mass as no geometry is given
the 30N and 40N forces will be at 30 deg to the horizontal and the verticle
be carefull about the signs of the directions !!!
hope this helps some
: )
Use tension X time = momentum exchange the time is the time of touch which you're given the momentum exchange would be (0.38x2.a million) - (0.38x-2.0) Then divide the momentum exchange (watch the indicators!) by making use of the touch time. solid luck