vector question?

Question

Two forces have magnitudes of 50 N and 29 N and the angle between them is 67°.

I would like to find the magnitude of the resultant in newtons and the angle it makes with the larger of the two forces in degrees

Any help would be greatly appreciated.

Answer 1

Add the vectors:
<50, 0> + <29·cos 67°, 29·sin 67°>
= <50, 0> + <29(0.3907), 29(0.9205)>
= <50, 0> + <11.3312, 26.69464>
= <61.3312, 26.69464>

The magnitude is:
√(61.3312² + 26.69464²)
= √(3761.516 + 712.6038)
= √4474.1198
= 66.889 Newtons

The angle is:
Atan (26.69464 / 61.3312)
= Atan (0.435254)
= 23.52°

Answer 2

OK. Since it hasn't been specified we may, without loss of generality, assume that the 50N force (lhe larger of the two) is at an angle of 0° and the 29N force is at an angle of 67°. Call the larger force A and the smaller B and break them into x and y components as follows:
Ax = 50cos(0°) = 50
Ay = 50sin(0°) = 0
Bx = 29cos(67°) = 11.3312 (to 4 decimal places)
By = 29sin(67°) = 26.6946 (to 4 decimal places)

Now add up the x and y components to get the C x and y components
Ax + Bx = Cx = 61.3312
Ay + By = Cy = 26.6946

Now convert back to polar coordinates (r,Θ) as follows
r = √(x² + x²) = √(61.3312² + 26.6949²) = 66.8890
Θ = arctan(y/x) = arctan(26.6946/61.3312) = 23.5212°

So the magnitude of the resultant force is about 66.9 N at an angle of about 23.5° to the larger force.

Count your blessings that you have a calculator to do the arithmetic. When I learned this (back in the late 50's) I had to use a 'slide rule' and do the additions with pencil and paper☺

Most scientific calculators have a single button that will do rectangular (x,y) coordinate to polar (r,Φ) coordinate and back transforms. If yours doesn't, think about investing 20 bucks or so for a K-mart or Walgreens (or whatever) special. It'll save you hours of time ☺


Doug

DAMN!!! That louise is both smart **and** quick ☺

Answer 3

Vector problems can always be solved by graphing them. First you lay out the 50 magnitude horizontally on a piece of graph paper, place the arrow head on the right end. At the arrow head measure an angle of 67 degrees, up or down, and then measure 29 units along this line and place an arrow head. The resultant force is the length of the line from the first vectors start point to the arrow head of the second vector. This procedure is called plotting vectors "tail to head."

Answer 4

First lock the first angle into a position, say at the origon with the force directed east.

Then break the two forces into x and y forces.
x1 = 50 * cos(0) = 50
y1 = 50 * sin(0) = 0

x2 = 29 * cos(67) = 11.33120
y2 = 29 * sin(67) = 26.69464

Now add the two together
x3 = 61.33120
y3 = 26.69464

Now use the arc tanget to get the angle.
tan(theta) = y3 / x3
tan(theta) = 26.69464 / 61.33120
theta = 23.521269 degrees

Answer 5

Let magnatiude=R
and Q & P are forces.
Also the angle between Q & P is x.
then, R=( P^2+Q^2+2PQcos(x) )^(1/2)
here, P=50N, Q=29N, cos(x)=67 degrees.
Then, R=(50^2+29^2+2*50*29*cos67)^(1/2) N
R=(2500+841+2900*0.3907)^(1/2) N
R=(3341+1133.03)^(1/2) N
R=(4474.03)^(1/2) N
R=66.8882N
========
The angle it makes with the larger of the two forces in degree=y
Then,
tan (y)=sin(67)/(50+cos(67))
tan(y)=0.9205/(50+0.3907)
=0.9205/50.3907
y =tan-1( 0.0183) degree
=====================

Answer 6

find the net forces of each force in horizontal and vertical direction. i.e. substract the horizontal force of 29N from the horizontal force of 50N. repeat for vertical.

Now, you have the net horizontal and vertical forces. calculate the resulting direction of the two.

Answer 7

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Answer 8

66.888N in a 23.52 degree direction from the larger.

You do sig figs.