Vector Question (magnitude of the resultant and the angle it makes with the larger of the two forces)?

Question

Two forces have magnitudes of 26 N and 6 N and the angle between them is 175°. Find the magnitude of the resultant and the angle it makes with the larger of the two forces to one decimal place

Any help would be gratefully appreciated as i am stuck on this question.

Answer 1

Hmm..

There is a little problem since you did not specify the direction of any of the forces except the angle between them. We can assume that the larger force of 26N is pointing in the direction of positive x-axis or say westward and then the other 6N force is pointing mostly eastward and just a little bit to the north. Also we can say that the algebraic sum of x and y components of our forces should yield a resultant. Then we have:
F1(x)=26
F1(y)=0
F2(x)= 6 cos (175)= - 5.98N
F2(y)= 6 sin (175)= .52N

Ft(x)= 20.02N
Ft(y)= .52N
Magnitude or the resultant |Ft|=sqrt((20.02)^2+(.52)^2) ~20

Angle = arctan (.52/20.02)= 1.49 degrees

That is it. Have fun

Answer 2

there is a general formula to find the magnitude

(resultant force)^2 = a^2 + b^2 - 2abcos(angle between the vectors) where a and b are the magnitude of the given vectors.

let r = the magnitude of the resultant force.

now we want to find the angle that r makes with the larger of the two forces (a=26N the larger one)

in vector notation: vector r = vector a + vector b
or
vector b = vector r - vector a
the above is a vector notation but there is a magnitude notation for this one, too but it differes from the resultant vector

b^2 = r^2 + a^2 + 2racos(angle between r and a) [note there is a plus sign not minus]

use this formula to find the angle between r and a. (a, b and r are the magnitude of the vectors a,b and r respectively!)

*
*
*
*

Note: i'm not sure about the second formula. maybe the right one is: b^2 = r^2 + a^2 - 2racos(angle between r and a) [there is a minus sign not plus]

check both of them.

Answer 3

Assume the 26 N force lies on the positive x axis. The 5 N force is at 175 degrees.

26 N cos(0) + 5 N cos (175)
25 N sin (0) + 5 N sin (175)

26 i - 4.98 i = 21.02 i
0j + 0.436 j

inv tan (0.436/21.02) = 1.2 degrees
((0.435)^2+(21.02)^2)^0.5 =21.02N

21.02 N at 1.2 degrees.

Answer 4

Solve this question by Parellogram Law of Vector. It will definetely give u the right answer