A crate weighs 820N and rests on a 34degree incline. You can keep it from sliding down the incline by pushing on it with a force of 240N applied parallel to the incline.
1)Calculate the coefficient of static friction between the crate and the incline.
2)What is the greatest force you can apply before the crate begins to move up the incline.
To solve both of these problems, we need to use the concept of static friction.
1) The coefficient of static friction can be calculated using the equation:
μs = (Ff / N)
Where:
μs is the coefficient of static friction
Ff is the force of friction
N is the normal force
Since the crate is on an inclined plane, we need to resolve the forces acting on it. The force of gravity acting on the crate can be divided into two components: one perpendicular to the incline (called the normal force) and one parallel to the incline (called the weight component parallel to the incline).
The weight component parallel to the incline can be calculated using the formula:
Wparallel = m * g * sin(θ)
Where:
m is the mass of the crate
g is the acceleration due to gravity (approximately 9.8 m/s^2)
θ is the angle of the incline (34 degrees)
Converting the weight in newtons to mass, we can calculate:
m = Wparallel / g
Next, we can calculate the normal force acting on the crate:
N = m * g * cos(θ)
Now we have all the values needed to calculate the coefficient of static friction:
μs = (Ff / N) = (240N / N) = 240 / (m * g * cos(θ))
Substituting the values, we can find the coefficient of static friction.
2) To determine the maximum force before the crate begins to move up the incline, we need to consider the forces acting on the crate. The two main forces are the force applied parallel to the incline and the force of friction acting in the opposite direction.
Since the crate is not moving, the force of friction is equal to the force applied parallel to the incline. Thus:
Ff = Force applied parallel to the incline
Substituting the values:
Ff = 240N
So, the greatest force that can be applied before the crate begins to move up the incline is 240N.