SOLUTION: I'm having trouble with the following example problem and would appreciate if I could get a walkthrough of how to do it. Forces with magnitudes of v = 160 newtons and u = 280 n

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Question 1206729: I'm having trouble with the following example problem and would appreciate if I could get a walkthrough of how to do it.
Forces with magnitudes of v = 160 newtons and u = 280 newtons act on a hook. The angle between the two forces is 45°. Find the magnitude of the resultant of this force.
Thank you.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website!
.
I'm having trouble with the following example problem and would appreciate if I could get a walkthrough
of how to do it.
Forces with magnitudes of v = 160 newtons and u = 280 newtons act on a hook.
The angle between the two forces is 45°. Find the magnitude of the resultant of this force.
Thank you.
~~~~~~~~~~~~~~~~~~~~~~~~~

        Do it exactly in accordance with the rule.

The rule of adding forces (vectors) is the parallelogram rule. In this case you have one vector with the magnitude 160 N and another vector with the magnitude 280 N. When you apply the parallelogram rule, the angle between the vectors is 135°. To find the resultant, use the Law of Cosine R^2 = 160^2 + 280^2 - 2*160*280*cos(135°) = = 167356.7676. Hence, R = = 409.1 N (rounded).

Solved.

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For introduction to vectors on a plane, see the lessons
    - Vectors in a plane
    - Sum of vectors that are coherently oriented sides of a convex closed polygon
    - Sum of vectors that are coherently oriented sides of an unclosed polygon
    - Sum of vectors that connect the center of a parallelogram with its vertices
    - Vectors in a coordinate plane
    - Addition, Subtraction and Multiplication by a number of vectors in a coordinate plane
    - Summing vectors that are coherently oriented sides of a convex closed polygon
    - Summing vectors that are coherently oriented sides of an unclosed polygon
    - The Centroid of a triangle is the Intersection point of its medians
    - The Centroid of a parallelogram is the Intersection point of its diagonals
    - Sum of vectors connecting the center of mass of a triangle with its vertices
    - Sum of vectors connecting the center of mass of a quadrilateral with its vertices
    - Sum of vectors connecting the center of mass of a n-sided polygon with its vertices
    - Sum of vectors connecting the center of a regular n-sided polygon with its vertices
    - Solved problems on vectors in a plane
    - Solved problems on vectors in a coordinate plane
    - HOW TO find the length of the vector in a coordinate plane
in this site.

Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

Let's say the 160 newton force is pulling directly eastward, and the other vector points in the northeast direction.

For the 160 newton vector
x = r*cos(theta) = 160*cos(0) = 160
y = r*sin(theta) = 160*sin(0) = 0
The < x,y > form of this vector is < 160,0 >

For the 280 newton vector
x = r*cos(theta) = 280*cos(45) = 280*sqrt(2)/2 = 140*sqrt(2)
y = r*sin(theta) = 280*sin(45) = 280*sqrt(2)/2 = 140*sqrt(2)
The < x,y > form of this vector is < 140*sqrt(2), 140*sqrt(2) >

The two vectors are
< 160,0 >
< 140*sqrt(2), 140*sqrt(2) >
Add straight down to get the resultant vector in component form.
< 160+140*sqrt(2), 140*sqrt(2) >

That approximates to roughly,
< 357.98989873, 197.98989873 >

The last set of steps is to compute the magnitude of the resultant.
r = magnitude = length of the vector
r = sqrt(x^2 + y^2) due to the Pythagorean theorem
r = sqrt( 357.98989873^2 + 197.98989873^2 )
r = 409.09261493 newtons
This answer is approximate. Round it however your teacher instructs.