The vector sum of the forces of 10N and 6N can be?

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Answer 1 · 2018-05-14T07:52:58

$the magnitudes of the vector sum can take values from 4 N to 16 N.$ $"the magnitudes of the vector sum can take values from 4 N to 16 N."$

$you can find the vector sum of two vectors using the formula below.$ $"you can find the vector sum of two vectors using the formula below."$

$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 \cdot F_{1} \cdot F_{2} \cdot cos (θ)}$ $R=sqrt(F_1^2+F_2^2+2*F_1*F_2*cos (theta))$

In this formula, R is the size of the total vector, $F_{1} and F_{2}$ $F_1 " and " F_2$ are the magnitudes of the summed vectors, and theta is the angle between the two vectors.
$θ$ $theta" "$ angle can take values from 0 to 180.
if $θ = 0 cos (θ) = 1$ $theta" "=0 " "cos (theta)=1$
$R = \sqrt{10^{2} + 6^{2} + 2 \cdot 10 \cdot 6} = \sqrt{100 + 36 + 120} = \sqrt{256} = 16 N$ $R=sqrt(10^2+6^2+2*10*6)=sqrt(100+36+120)=sqrt(256)=16N$
if $θ = 180 cos (θ) = - 1$ $theta" "=180 " "cos (theta)=-1$
$R = \sqrt{10^{2} + 6^{2} - 2 \cdot 10 \cdot 6} = \sqrt{100 + 36 - 120} = \sqrt{16} = 4 N$ $R=sqrt(10^2+6^2-2*10*6)=sqrt(100+36-120)=sqrt(16)=4N$
the magnitudes of the vector sum can take values from 4 N to 16 N.
$4 \leq R \leq 16$ $4<=R<=16$