A 20-kg crate is lowered down a plane inclined at an angle of 37 with the horizontal by a rope that exerts on the crate a 50-n force directed parallel to the plane. A 40-n friction force, also parallel to the plane, opposes the downward motion. What is the acceleration of the crate?
Please help me answer these
To find the acceleration of the crate, we need to consider the forces acting on it.
Let's break down the forces acting on the crate along the incline:
1. Force parallel to the plane (Fpar): This force is exerted by the rope and directed downward, parallel to the incline. Its magnitude is 50 N.
2. Force of friction (Ffr): This force opposes the motion of the crate and is directed upward, parallel to the incline. Its magnitude is 40 N.
3. Component of gravity parallel to the incline (Fgpar): This force is due to the weight of the crate and acts parallel to the incline. Its magnitude can be calculated as follows:
Fgpar = m * g * sin(θ)
Where:
m = mass of the crate = 20 kg
g = acceleration due to gravity = 9.8 m/s²
θ = angle of the incline = 37°
Plugging in the values, we have:
Fgpar = 20 kg * 9.8 m/s² * sin(37°)
4. Net force (Fnet): This is the sum of the forces acting along the incline, given by:
Fnet = Fpar - Ffr - Fgpar
Now, we can calculate the acceleration (a) using Newton's second law:
Fnet = m * a
Rearranging the equation, we get:
a = Fnet / m
To find the acceleration, we can substitute the values into the equation and solve for a.