find the coefficient of friction....please help!!?

Question

a horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin 7/25 to the horizontal. Find the coefficient of friction between the block and the plane.
and the acceleration with which the block will move when the force is removed.
pleasse help ...

Answer 1

Fun with geometry!!

I'll provide the method, you can plug in the numbers

theta = sin^-1(7/25)

sin(theta) = 7/25

sin(theta) = opposite/hypotenuse

Using pythagorean theorem we know that the adjacent side it 24.

cos(theta) = 24/25

Back to the problem...

Draw a FBD of the block

Sum of the forces along the plane = 0 = F - friction - m*g*sin(theta)

Sum of the forces normal to the plane = 0 = N - m*g*cos(theta)

N = m*g*cos(theta)

friction is defined as friction = u*N

Insert that back into the top equation

F - friction - m*g*sin(theta) = 0
F - u*N - m*g*sin(theta) = 0
F - u*m*g*cos(theta) - m*g*sin(theta) = 0

solve for u

u = (F - m*g*sin(theta)) / (m*g*cos(theta))

Presto!

To find the acceleration, take the same equation and instead of the sum of the forces along the plane equalling 0, it equals m*a

m*a = - friction - m*g*sin(theta)
a = (-u*m*g*cos(theta) - m*g*sin(theta))/m
a = -u*g*cos(theta) - g*sin(theta)