Rather than rotate completely horizontally, a carousel rotates at an angle of 10° from horizontal, with a constant angular velocity of 1.571 RAD/s. A box is placed 1.5 m from the axis of rotation. What’s the minimum coefficient of static friction so the box will not slip?
Well, I must say, this carousel seems to be quite the angle enthusiast! Now, let's crunch some numbers to find the minimum coefficient of static friction so our box can hold on tight.
First, let's calculate the acceleration of the box towards the center of the carousel. We know that the acceleration (a) is given by:
a = r * ω^2
where r is the distance from the axis of rotation (which is 1.5 m in this case) and ω is the angular velocity (1.571 RAD/s).
Plugging in the values, we get:
a = 1.5 * 1.571^2
Approximately, this gives us an acceleration of 3.694 m/s^2.
Now, to find the minimum coefficient of static friction (μ), we can use the equation:
μ ≥ tan(θ)
where θ is the angle of inclination (10° in this case).
Plugging in the values, we get:
μ ≥ tan(10°)
Using a calculator, this gives us:
μ ≥ 0.176
So, the minimum coefficient of static friction required to prevent the box from slipping is approximately 0.176. We sure hope the box has some grippy shoes on!
To find the minimum coefficient of static friction so that the box will not slip on the rotating carousel, we need to consider the forces acting on the box.
First, let's draw a free-body diagram of the forces acting on the box:
1. Weight (mg): This force acts vertically downwards and is equal to the mass (m) of the box multiplied by the acceleration due to gravity (g).
2. Normal force (N): This force acts perpendicular to the surface of contact between the box and the carousel. It counteracts the weight of the box and prevents it from sinking into the carousel.
3. Static friction force (f_s): This force acts parallel to the surface of contact between the box and the carousel. It prevents the box from slipping on the rotating carousel.
The angle of rotation of the carousel (θ) is given as 10° from horizontal. We'll need to resolve the weight force parallel and perpendicular to the surface to consider its effect on the friction force.
Now, let's analyze the forces in the vertical direction:
Summing the forces in the vertical direction, we have:
N - mgcosθ = 0,
where mgcosθ is the vertical component of the weight force.
Since the box is not sinking into the carousel, the normal force (N) is equal to the vertical component of the weight force:
N = mgcosθ.
Now, let's consider the forces in the horizontal direction:
Summing the forces in the horizontal direction, we have:
f_s - mgsinθ = ma,
where mgsinθ is the horizontal component of the weight force and a is the centripetal acceleration.
The centripetal acceleration (a) can be found using the relationship between angular velocity (ω) and tangential velocity (v):
a = rω²,
where r is the distance from the axis of rotation to the box.
In this case, r = 1.5 m, and the angular velocity (ω) is given as 1.571 rad/s. So, we can calculate the centripetal acceleration (a):
a = (1.5 m)(1.571 rad/s)².
Then, substitute this value of a into the equation:
f_s - mgsinθ = m(1.5 m)(1.571 rad/s)².
Finally, rearrange the equation to solve for the minimum coefficient of static friction (μ_s):
f_s = μ_sN.
Substituting the value of N from above and rearranging the equation, we have:
μ_s(mgcosθ) = m(1.5 m)(1.571 rad/s)² + mgsinθ.
Divide both sides by mgcosθ:
μ_s = [(1.5 m)(1.571 rad/s)² + gsinθ] / gcosθ.
Now, plug in the known values: g = 9.8 m/s² and θ = 10° (convert to radians):
μ_s = [(1.5 m)(1.571 rad/s)² + (9.8 m/s²)(sin(10°))] / (9.8 m/s²)(cos(10°)).
Calculating this expression will give you the minimum coefficient of static friction required for the box to not slip on the rotating carousel.