Question
Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
✓
Answer
From the free body diagram$\text{g}=10\text{m/s}^2,\text{m}=2\text{kg},\theta=30^{\circ},\mu=0.2$
$\text{R}-\text{mg}\cos\theta-\text{F}\sin\theta=0$
$\Rightarrow\text{R = mg}\cos\theta+\text{F}\sin\theta \ ...(1)$
And $\text{mg}\sin\theta+\mu\text{R}-\text{F}\cos\theta=0$
$\Rightarrow\text{mg}\sin\theta+\mu(\text{mg}\cos\theta+\text{F}\sin\theta)-\text{F}\cos\theta=0$
$\Rightarrow\text{mg}\sin\theta+\mu\text{ mg}\cos\theta+\mu\text{F}\sin\theta-\text{F}\cos\theta=0$
$\Rightarrow\text{F}=\frac{(\text{mg}\sin\theta+\mu\text{mg}\cos\theta)}{(\mu\sin\theta-\cos\theta)}$
$\Rightarrow\text{F}=\frac{2\times10\times\big(\frac{1}{2}\big)+0.2\times2\times10\times\big(\frac{\sqrt{3}}{2}\big)}{0.2\times\big(\frac{1}{2}\big)-\big(\frac{\sqrt{3}}{2}\big)}$
$=\frac{13.464}{0.76}=17.7\text{N}\approx17.5\text{N}$
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