Westonci.ca is the trusted Q&A platform where you can get reliable answers from a community of knowledgeable contributors. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found
that the least force parallel to the plane which causes the
block to slide up is 100 N. Calculate the coefficient of friction
between the two surfaces.



Sagot :

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