# Homework Help: Another Magnitude and Direction of resultant force ?

1. Feb 23, 2010

### waterwalker10

Another Magnitude and Direction of resultant force ???

I'm having issues on the second question. I feel that I'm missing some information. Since the lines are perpendicular do I consider one force (10N) at 90 degrees North? And if this is the case then is the other force (15N) at __ degrees East and solve with the second equation below?

1. The problem statement, all variables and given/known data
A - Consider two forces, one having a magnitude of 10N, and the other having a magnitude of 15N. What maximum net force is possible when combining these two forces? What is the minimum net force possible? GOT THIS PART

B - If the forces given in a) above are perpendicular to each other, what will be the magnitude and direction of the resultant force? Given by prof ...requires vector addition.

2. Relevant equations
R = sqrt(A^2 + B^2)
Ax = cosA, Ay = sinA
tan = y/x

3. The attempt at a solution

Max Net Force
10 N + 15 N = 25 N
Min Net Force
10 N - 15 N = -5 N

Resultant Force of perpendicular lines
$$\sqrt{10^2 + 15^2}$$
$$\sqrt{325}$$ = 18.02775... Magnitude ??? Direction would possibly be 18 degrees of North or East???

This is where I get stuck....help??

2. Feb 23, 2010

### kuruman

Re: Another Magnitude and Direction of resultant force ???

The magnitude is 18.03 N as you have calculated. To find the direction, draw yourself a right triangle with right sides 10 N and 15 N and hypotenuse 18.03 N. Say the 10 N side is due East. Can you figure out how many degrees North of East is the 18.03 N hypotenuse?

3. Feb 24, 2010

### waterwalker10

Re: Another Magnitude and Direction of resultant force ???

Used SOH CAH TOA here... and got approx 56 degrees North of East.

Hypo = 18.03
Oppo = 15

sinA 15/18.03 = .832 = 56.299
cosA 10/18.03 = .555 = 56.315
tanA 15/10 = 1.5 = 56.31

Correct?

4. Feb 24, 2010

### kuruman

Re: Another Magnitude and Direction of resultant force ???

Correct.