A block is at rest on the incline shown in the
figure. The coefficients of static and kinetic
friction are µs = 0.47 and µk = 0.4, respectively.
The acceleration of gravity is 9.8 m/s^2
The block is 44 kg
The angle is 22 degrees
What is the frictional force acting on the
44 kg mass?
I already have the frictional force which is 431.2
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Answer in units of ◦
What is the acceleration of the block down
the incline if the angle of the incline is 30 ◦?
Answer in units of m/s^2
normal force = m g cos A
so friction force up slope = mu m g cos A
weight component down slope = m g sin A
so mg (sin A - mu cos A ) = m a
==================================
at A = 22 , m = 44 , mu = .47
mu m g cos A = .47 * 44 * 9.8 * 0.927 = 188 N
Hmmmm
===================================
for a = acceleration = 0
sin A = mu cos A
tan A = mu = 0.47
A = 25.1 degrees
whew, close to slipping
==================================
mg (sin A - mu cos A ) = m a
a / g = sin 30 - 0.40 cos 30
=.5 - 0.346
a = 9.8 * 0.154 = 1.5 m/s^2