A) 14.14 N
B) 17.32 N
C) 10 N
D) 16.66 N
E) 0.866 N
Correct Answer: B
Solution :
The component of weight mg of block along the inclined plane\[=mg\text{ }sin\,\theta \]. The minimum frictional force to be overcome is also\[mg\text{ }sin\] \[\theta \]. To make the block just move up the plane the minimum force applied must overcome the component\[mg\text{ }sin\,\theta \]of gravitational force as well as the frictional force\[mg\text{ }sin\,\theta =2mg\] \[\sin \theta \]. \[\therefore \] Force needed\[=2mg\text{ }sin\theta \] \[=2\times 1\times 10\times \sin {{60}^{o}}\] \[=20\times \frac{\sqrt{3}}{2}\] \[=17.32\text{ }N\]
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