Question

A rectangular of mass 10kg rest on an inclined plane with a coefficient of static friction of 0.55 and coefficient of kinetic friction 0.25.calculate angle that the box will begin to sliding and it's acceleration

Answer 1

I need answer

Answer 2

weight normal to plane = m g cos A
so static friction up slope = 0.55 m g cos A
weight component down slope = m g sin A
starts to slide when
0.55 m g cos A = m g sin A
or
tan A = .55
A = 28.8 degrees

Answer 3

net force down slope when sliding = m g sin 28.8 - .25 m g cos 28.8 = m a
so
a = g (sin 28.8 - .25 cos 28.8)

note: who cares what the mass was?

Answer 4

I could not understand the solution

Answer 5

the first person who did this question I did not understand it

Answer 6

What a hell? I was looking for solution not explanations

Answer 7

Mk=0.25
Ms=0.55
Mass=10kg
For the mass begin to slide:
Tan -1 0.25=angle
Angle=14.04°

F=M×a
F=Mk×mgCos 14.04°
Ma=Mk×mgCos. 14.04°
Mass ❌ is cancelled
a=0.25×10Cos14.04°
a=2.4m/s^2