Represent on a diagram the following to vectors

  • Thread starter Thread starter undertow2005
  • Start date Start date
  • Tags Tags
    Diagram Vectors
AI Thread Summary
The discussion focuses on resolving two vectors: one measuring 3m at 45 degrees north of east and another measuring 6m at 60 degrees west of north. The initial calculations for the components of the vectors are critiqued, particularly the need to adjust the second vector's angle to reference east instead of north. The correct approach involves converting the angle of the second vector to 150 degrees to accurately determine its components. The importance of understanding the principle directions and quadrant signs for sine and cosine functions is emphasized for accurate resultant calculations. The conversation concludes with a reminder to visualize the vectors on a graph for clarity in determining their directions.
undertow2005
heres the problem

represent on a diagram the following to vectors:1) 3m, 45 degrees north of east; and 2) 6m, 60 deg west of north. find the resultant of sum of these two vectorsin terms of magnitude and angle with respect to the east.


nevermind the first part of the Q, i need to find the second part


my incorrect solution:

for vector one:
Cx=3m*cos45=2.121
Cy=3*sin45=2.121

Vector 1=2.121i + 2.121j

Vector two:
Cx=6*cos60=3
Cy=6*sin60=5.196

vector 2= 3i +5.196

i added them and got this 5.121i +7.317j

someone check!

My Final only two hours away!
 
Physics news on Phys.org
You've got the mechanics of it down, but you're not paying enough attention to your principle directions. These vectors aren't referenced to the same principle direction.

The first is referenced to East - your components were correct.

The second is referenced to North - you need to add 90 degrees to the angle to convert it to an East reference (you also made a mistake by using positive 60, which is 60 degrees East of North, but that's irrelevant since you started from the wrong reference).

The 'i' of your second vector is the cosine of 150 degrees - the 'j' the sine of 150 degrees. (the cosine of 150 is negative, meaning your resultant will be a positive plus a negative).
 
BobG said:
You've got the mechanics of it down, but you're not paying enough attention to your principle directions. These vectors aren't referenced to the same principle direction.

The first is referenced to East - your components were correct.

The second is referenced to North - you need to add 90 degrees to the angle to convert it to an East reference (you also made a mistake by using positive 60, which is 60 degrees East of North, but that's irrelevant since you started from the wrong reference).

The 'i' of your second vector is the cosine of 150 degrees - the 'j' the sine of 150 degrees. (the cosine of 150 is negative, meaning your resultant will be a positive plus a negative).

vector 2

6*cos(-150=-5.196
6*sin(-150=-3.00

correct?
 
Almost. Sine of 150 is positive one-half (+0.5).

Sine is positive from 0 to 180

Cosine is positive from 270 to 90

If you plot this on a graph, your cosine is your x and your sine is your y, which makes it pretty easy to see which quadrants are positive or negative.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top