a block that has a mass of 36 kg is initially at rest on an 20 degree inclined plane. (a) find the component of the weight that is parallel to the plane. (b) find the component of the weight that is perpendicular to the plane. (c) find the maximum force of static friction if the coefficient of static friction is .29 (d) find the acceleration of the block if the coefficient of kinetic friction is .18.
2007-02-20 12:51:31 · 2 answers · asked by argentina_mandy20 1 in Science & Mathematics ➔ Physics
a) Parallel = m*g*sin(20)
b) Perpendicular =m*g*cos(20)
c) the force of static friction is the parallel component
if the parallel component exceeds the maximum static friction, there is an acceleration working against a kinetic frictional force.
The way the question is worded in c, I will assume that the angle is variable
so the maximum angle of incline before breaking static friction
m*g*sin(th)=m*g*cos(th)*u
th = angle of incline
u= coefficient of friction
tan(th)=u
for a u=.29
th = 16.17 degrees
which means at 20 degrees, the block is accelerating
d) the acceleration is
m*a=m*g*sin(20)-m*g*cos(20)*.18
mass divides out
a=g*(sin(20)-cos(20)*.18)
j
2007-02-20 12:59:04 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
the weight that is taken parrellel to the ramp is Fweight(x)
the wieght that is taken perpendicular to the ramp is Fweight(y)
Fweight(x) = sin(20)36*9.8 = 120.66N
Fweight(y) = cos(20) 36*9.8 = 331.52N
Friction = u(Fnormal)
Fnormal is equal to Fweight(y)
Friction = .29(331.52)
Friciton = 96.14N
a = Fnet/mass
since Fweigh(x) and Friction make up Fnet, we have:
a = (Fweigh(x) - Friciton) / mass
a = (120.66 - 96.14)/36
a = .681m/s^2
hope this helps
2007-02-20 21:04:55 · answer #2 · answered by 7 · 0⤊ 0⤋