A block is at rest on the incline shown in the
figure. The coefficients of static and kinetic
friction are µs = 0.4 and µk = 0.34, respectively.
The acceleration of gravity is 9.8 m/s.
What is the largest angle which the incline
can have so that the mass does not slide down
the incline?
Answer in units of degrees
normal force = m g cos T
force holding block = .4 m g cos T
force of gravity down slope = m g sin T
so
.4 cos T = sin T
so
tan T = .4
T = 21.8 degrees
To find the largest angle at which the mass does not slide down the incline, we need to compare the force of gravity pulling the mass down the incline to the maximum static friction force that can oppose it.
First, let's calculate the maximum static friction force. The maximum static friction force (Fsf) can be found using the formula:
Fsf = µs * N
where µs is the coefficient of static friction and N is the normal force.
The normal force (N) is equal to the sum of the component of the weight perpendicular to the incline.
N = mg * cosθ
where m is the mass and g is the acceleration due to gravity.
Using trigonometry, the component of the weight perpendicular to the incline can be expressed as:
mg * cosθ = mg * sin(90° - θ) = mg * sinθ
Therefore, the normal force becomes:
N = mg * sinθ
Substituting the expression for N back into the equation for maximum static friction force, we get:
Fsf = µs * mg * sinθ
Comparing the maximum static friction force to the force of gravity (mg), we have:
µs * mg * sinθ = mg
Simplifying the equation, we have:
µs * sinθ = 1
Now we can solve for the angle θ:
sinθ = 1 / µs
θ = arcsin(1 / µs)
Plugging in the value of the coefficient of static friction (µs = 0.4), we can calculate the angle:
θ = arcsin(1 / 0.4)
Using a calculator or a trigonometric table, we find:
θ ≈ 24.6 degrees
So, the largest angle at which the mass does not slide down the incline is approximately 24.6 degrees.
To determine the largest angle at which the mass does not slide down the incline, we need to consider the forces acting on the block.
First, let's consider the forces acting perpendicular to the incline. The weight of the block acts straight down and can be broken into two components: mg * cos(theta) acts perpendicular to the incline, and mg * sin(theta) acts parallel to the incline.
Next, let's consider the forces acting parallel to the incline. We have the force of static friction (Fs) and the force of gravity acting down the incline.
For the block to remain at rest, the force of static friction (Fs) must be equal to or greater than the force of gravity acting down the incline (mg * sin(theta)). The formula for static friction is Fs = µs * (mg * cos(theta)), where µs is the coefficient of static friction.
So, we can set up the following equation:
µs * (mg * cos(theta)) >= mg * sin(theta)
Now, let's solve for theta, the angle of the incline, to find at what angle the block will not slide down the incline.
µs * (mg * cos(theta)) >= mg * sin(theta)
Divide both sides of the equation by mg to cancel out the mass:
µs * cos(theta) >= sin(theta)
Divide both sides by cos(theta):
µs >= tan(theta)
Taking the arctan (inverse tan) of both sides:
theta <= arctan(µs)
Now, let's substitute the given value of µs = 0.4 into the equation:
theta <= arctan(0.4)
Using a calculator to find the arctan(0.4), we get approximately 21.80 degrees.
Therefore, the largest angle the incline can have so that the mass does not slide down is approximately 21.80 degrees.