If by theta you mean the angle at the base of slope on which is the body laying, and you want to calculate minimal theta for which the blocks starts to slide:
Let's first calculate:
weight: Q = mg
force normal to the slope: N = Q cos theta = mg cos theta
force tangent to the slope: F = Q sin theta = mg sin theta
force of friction: T = fN = fmg cos theta, where f is coefficient of friction
The body will start to move downwards, when T = F, or:
fmg cos theta = mg sin theta
which after simplyfying becomes:
f cos theta = sin theta,
f = sin theta / cos theta
f = tan theta
Therefore,
theta = arc tan f
As you see, the angle only depends on friction coefficient f.
(If that's not a problem you asked to be solved, edit your question please to precisely state what needs to be calculated)
The coefficient of friction is the tangent of the angle theta where the angle is measured from horizontal when the mass first starts to slip
T1/T2=e^(mu*theta)where T1/2 are the tensions in the circlemu is the coefficient of frictiontheta is the angle of the circle in contact with the rope.
You need to know both material involved in the friction to find the coefficient
This coefficient of static friction is needed to find the frictional force between a body and a surface on which body has to move. If u (mu) is the coefficient of friction then uR gives the frictional force between moving body and surface. There is no unit for coefficient of friction. Here R is reaction which equals to the weight of the body
Given the vector in angle-radius form? y-component=r sin(theta), x-component=r cos(theta)
There is no minimum value for the coefficient of friction. And the linear acceleration will depend on its unknown value.
The answer depends on what theta represents!
tan(theta) = 1 then theta = tan-1(1) + n*pi where n is an integer = pi/4 + n*pi or pi*(1/4 + n) Within the given range, this gives theta = pi/4 and 5*pi/4
Cotan(theta) is the reciprocal of the tan(theta). So, cot(theta) = 1/2.
No.
The equation for friction is F=uN. F (friction), u (coefficient of friction), and N (normal). So you first need to solve for the normal by using Newton's second law. Also solve for the x component of the gravity force. Since it is static friction, you know it should be at rest, so that x component force should be the same as the force of friction. Knowing that and the normal, plug it into the equation and solve for u.
Since theta is in the second quadrant, sin(theta) is positive. sin2(theta) = 1 - cos2(theta) = 0.803 So sin(theta) = +sqrt(0.803) = 0.896.