Another Magnitude and Direction of resultant force ?

[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?

Homework Statement

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.

Homework Equations

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

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??

 

[kuruman]

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?

 

[waterwalker10]

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?

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

Hypo = 18.03
Adja = 10
Oppo = 15

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

Correct?

 

[kuruman]

Correct.