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KATRIN_1 [288]
2 years ago
13

An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found

Physics
1 answer:
Kaylis [27]2 years ago
5 0

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒   µ = (100 N) / (84.9 N) ≈ 1.2

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The guy below is wrong!


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Using the known values in the formula, we get:

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The detailed solution can be found in the attachment below.

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