An object with a mass of $14 k g$ $14 kg$ is acted on by two forces. The first is $F_{1} = < 5 N, 6 N >$ $F_1= < 5 N , 6 N>$ and the second is $F_{2} = < 3 N, 7 N >$ $F_2 = < 3 N, 7 N>$ . What is the object's rate and direction of acceleration?

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Answer 1 · 2017-03-25T09:23:55

The rate of acceleration is $= 1.09 m s^{- 2}$ $=1.09ms^-2$ in the direction of $= 58.4$ $=58.4$ º

The resultant force is

$\to F = {\to F}_{1} + {\to F}_{2}$ $vecF=vecF_1+vecF_2$

$= < 5, 6 > + < 3, 7 >$ $= <5,6>+<3,7>$

$= < 8, 13 >$ $=<8,13>$

$\to F = m \to a$ $vecF=m veca$

$\to a = \frac{1}{m} \cdot \to F$ $veca =1/m*vecF$

$= \frac{1}{14} < 8, 13 > = < \frac{8}{14}, \frac{13}{14} >$ $=1/14<8,13> = <8/14, 13/14>$

The magnitude of the acceleration is

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ < \frac{8}{14}, \frac{13}{14} > ∣ ∣ ∣ ∣ ∣ ∣$ $||veca||=||<8/14,13/14>||$

$= \sqrt{{(\frac{8}{14})}^{2} + {(\frac{13}{14})}^{2}}$ $=sqrt((8/14)^2+(13/14)^2)$

$= \frac{\sqrt{233}}{14} = 1.09 m s^{- 2}$ $=sqrt(233)/14=1.09ms^-2$

The direction is $θ = arctan (\frac{13}{8})$ $theta = arctan((13)/(8))$

The angle is in the 1st quadrant

$θ = arctan (\frac{13}{8}) = 58.4$ $theta=arctan(13/8)=58.4$ º

An object with a mass of 14 kg14kg14 kg is acted on by two forces. The first is F_1= < 5 N , 6 N>F1=<5N,6N>F_1= < 5 N , 6 N> and the second is F_2 = < 3 N, 7 N>F2=<3N,7N>F_2 = < 3 N, 7 N>. What is the object's rate and direction of acceleration?