Question

If the co-efficient of static friction is 0.75, the angle through an inclined plane must be raised so
that the block placed on it may begin to slide is sin 37
5​

Answer 1

Answer:$37^{\circ}$

Explanation:

Given

Coefficient of static friction is $\mu =0.75$

Block will remain in equilibrium until

$\tan \theta \leq \mu$

At critical point

$\tan \theta =0.75$

$\theta =\tan ^{-1}(0.75)$

$\theta =36.86\approx 37^{\circ}$

after $\theta$ is greater than $37^{\circ}$ block begin to slide

Answer 2

Given

Coefficient of static friction is \mu =0.75μ=0.75

Block will remain in equilibrium until

\tan \theta \leq \mutanθ≤μ

At critical point

\tan \theta =0.75tanθ=0.75

\theta =\tan ^{-1}(0.75)θ=tan

−1

(0.75)

\theta =36.86\approx 37^{\circ}θ=36.86≈37

∘

after \thetaθ is greater than 37^{\circ}37

∘

block begin to slide