VECTORS help!!?

Question

Hey! I've been trying to work on this vectors worksheet lately. Here it is:

1. http://img205.imageshack.us/img205/9427/70466397yk4.jpg
2. http://img490.imageshack.us/img490/2572/58282826kh9.jpg
3. http://img490.imageshack.us/img490/8449/72000225yt9.jpg

I got many answers, but I was unable to do (or was not sure about) the ones with stars next to them (#4, 7, 10, 12, 14, and 20). I'd really appreciate if someone explained them to me. Also, if someone has the time, and heart to look over my other answers and tell me if I'm correct or not, I'd be more than grateful.

Thanks for any help! :]

Thank you SO much. Both of you are lifesavers.

Answer 1

In general a vector will have components in x, y, z and and any other number of dimensions. In this case we are dealing with x, y components.

So we can state that a vector in two dimensions is
V= Vx i + Vy j where

Vx = V cos(Theta)
Vy =V sin(Theta)

Now we are ready for #4

Since V= 10 and Vx=6
cos(Theta)=Vx/V=
Theta= arcCos(Vx/V)=
Theta= arcCos(6/10)=53.1 degrees

#7

Fx1= 50, Fx2=-50
Fy1= 50, Fy2=-50

F(resultant)=[Summation of x components]i + [Summation of y components]j

Summation of x components = Fx1+Fx2= 50 - 50=0

Summation of y components = Fy1+Fy2= 50 - 50=0

Then F(resultant)= oi+0j=0

#10
We have to decompose the forces into component

Sum of Fx = 100 x cos (60) - 80 cos(75)=
Sum of Fx =50-20.7 =29.3 N
Sum of Fy = 100 x sin (60) + 80 sin(75)=
Sum of Fy = 86.6 + 77.3=163.9N

F= Fx i +Fy j

tan(Theta)= Fy/Fx
Theta = arcTan(Fy/Fx)
Theta= arcTan(163.9/29.3)=79.9 deg

F= sqrt(Fx^2 + Fy^2)
F= sqrt(29.3^2 + 163.9^2)
F=167 N

#12 since you and the other car are traveling in the same direction the relative to you the speed is the difference between these vectors

V(relative) = V1-V2= 45 - 65= -20 mi/h

Notice the - sign and it makes sense since to you it would appear to move in a negative x direction.

#14
Again we use the idea that the sum of all forces is zero

Sum(Fy)= Fy1 + Fy2 - mg=0
(we dont have to vorry about Fx since we have only one unknown)
also since F1=F2 we have

mg= 2Fy=2Fsin(37)
m= 2 F sin(37)/g
m= 2 x 75 x sin(37)/9.81
m= 9.2kg

#20

This is funny question
We know that the force of friction f is
f= uN= u mg cos(Theta) and

So when we increase the angle the angle Theta the normal decreases and so is the friction force.

It is definitely (a)

Let me know if you have any further questions.