A block with a weight of 36 N is placed on a sheet of aluminum with an inclination of 18. The coefficients of static and kinetic friction between the block and aluminum are 0.35 and 0.23 respectively. Will the block slide down the plane?
A) No. the downward force along the plane is 3.9 N and the maximum static friction force is 34 N
B) yes. The downward force along the plane is 11 N and the maximum static friction force is 7.9 N
C) no. The downward force along the plane is 11 N and the maximum static friction force is 12 N
D) yes. The downward force along the plane is 34 N and the maximum static friction force is 2.6 N
normal force = m g cos 18 = 36 * 0.951 = 34.2 N
so max static friction = .35 * 34.2 = 12 N
component of weight down ramp = 36 sin 18 = 11.1 N
So - Ok unless we get an earthquake. Once it gets started the kinetic friction will not stop it.
how did you get 0.951
To determine if the block will slide down the plane, we need to compare the force of gravity pulling the block down the incline (the component of its weight parallel to the plane) to the maximum frictional force that can be exerted by the aluminum.
First, we need to find the component of the weight of the block that acts parallel to the incline. This can be calculated using the formula:
\(F_{\text{parallel}} = F_{\text{gravity}} \times \sin(\text{angle of inclination})\)
Given that the weight of the block is 36 N and the angle of inclination is 18, we can substitute the values:
\(F_{\text{parallel}} = 36 \times \sin(18)\)
Next, we need to find the maximum frictional force that can be exerted by the aluminum. This can be calculated using the formula:
\(F_{\text{friction, max}} = F_{\text{normal}} \times \mu_{\text{static}}\)
The normal force, \(F_{\text{normal}}\), is equal to the perpendicular component of the weight of the block acting on the incline and can be calculated using the formula:
\(F_{\text{normal}} = F_{\text{gravity}} \times \cos(\text{angle of inclination})\)
Substituting the given values, we get:
\(F_{\text{normal}} = 36 \times \cos(18)\)
Finally, substituting the calculated values into the equation for the maximum frictional force:
\(F_{\text{friction, max}} = (36 \times \cos(18)) \times 0.35\)
By calculating the values, we have:
\(F_{\text{parallel}} \approx 10.98 \, \text{N}\)
\(F_{\text{friction, max}} \approx 13.23 \, \text{N}\)
Since the force of gravity parallel to the incline (10.98 N) is less than the maximum frictional force (13.23 N), the block will not slide down the plane. It will remain stationary due to the static friction between the block and the aluminum surface.