Determine the magnitude of the resultant force acting on a 7.5 −kg particle at the instant t=2 s, if the particle is moving along a horizontal path defined by the equations r=(2t+10) m and θ=(1.5t2−6t) rad, where t is in seconds
Solution
$$ \begin{aligned} &\text { Mass }=m=7.5 \mathrm{~kg} \\ &\text { time }=t=2 \mathrm{~s} \\ &r=2 t+10 \end{aligned} $$
differentiating
$$ \begin{aligned} &V_{r}=2 \\ &a r=0 \mathrm{~m} / \mathrm{s}^{2} \\ &\theta=1.5 t^{2}-6 t \end{aligned} $$
differentiating
$$ w=2(1.5 t-6) $$
again differentiating
$$ \begin{aligned} &\alpha=2(1.5) \mathrm{rad} / \mathrm{s}^{2} \\ &\alpha=3 \mathrm{rad} / \mathrm{s}^{2} \end{aligned} $$
Now finding \(a \theta=r \times \alpha\)
$$ \begin{aligned} &=14 \times 3 \\ &=42 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$
Now the componets of force are
$$ \begin{aligned} &\mathrm{Fr}=\mathrm{m} \mathrm{ar}=\mathrm{ON} \\ &\mathrm{F} \theta=\operatorname{ma} \theta=7.5 \mathrm{~kg}\left(42 \mathrm{~m} / \mathrm{s}^{2}\right) \\ &\mathrm{F} \theta=315 \mathrm{~N} \end{aligned} $$
The magnitude of resultant force would be
$$ \begin{aligned} &F=\sqrt{F r^{2}+F \theta^{2}} \\ &F=\sqrt{315^{2}} \\ &F=215 \mathrm{~N} \end{aligned} $$
This will be the magnitude of resultant force.
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