Register to reply

Another Magnitude and Direction of resultant force ?

by waterwalker10
Tags: direction, magnitude, resultant vectors
Share this thread:
waterwalker10
#1
Feb23-10, 09:10 PM
P: 16
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
[tex]\sqrt{10^2 + 15^2}[/tex]
[tex]\sqrt{325}[/tex] = 18.02775... Magnitude ??? Direction would possibly be 18 degrees of North or East???

This is where I get stuck....help??
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
kuruman
#2
Feb23-10, 09:30 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,442
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
#3
Feb24-10, 02:41 PM
P: 16
Quote Quote by kuruman View Post
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
#4
Feb24-10, 05:11 PM
HW Helper
PF Gold
kuruman's Avatar
P: 3,442
Another Magnitude and Direction of resultant force ?

Correct.


Register to reply

Related Discussions
Finding the Magnitude and Direction of Force one with the Resultant Given Introductory Physics Homework 15
Cosine and Sine rules to get magnitude and direction of a resultant force Introductory Physics Homework 4
Calculate the magnitude, direction and sense of the resultant force Engineering, Comp Sci, & Technology Homework 1
Finding magnitude and direction of resultant force Introductory Physics Homework 5
Magnitude and direction of the resultant. Introductory Physics Homework 1