an inclined plane rises to aheight of 2 meter over a distance of 6 meter.
calculate,
1, the angle of the slope
2, the velocity ratio(ideal mechanical advantage) of inclined plane
3, the theocratical force required to push an object with a mass of 200 Kg up the slope
1. The angle of the slope
To find the angle of the slope, we can use the formula for the angle of inclination:
angle = arctan(opposite/adjacent)
Where opposite is the vertical height (2 meters) and adjacent is the horizontal distance (6 meters).
angle = arctan(2/6)
angle = arctan(0.3333)
angle ≈ 18.4°
So, the angle of the slope is approximately 18.4 degrees.
2. The velocity ratio (ideal mechanical advantage) of the inclined plane
The ideal mechanical advantage (IMA) of an inclined plane is the ratio of the length of the slope to the height of the slope.
IMA = length of slope / height of slope
First, we need to find the length of the slope. We can use the Pythagorean theorem for that:
slope_length² = height² + distance²
slope_length² = 2² + 6²
slope_length² = 4 + 36
slope_length² = 40
slope_length = √40 ≈ 6.32 meters
Now we can find the IMA:
IMA = 6.32 / 2
IMA ≈ 3.16
So, the ideal mechanical advantage of the inclined plane is approximately 3.16.
3. The theoretical force required to push an object with a mass of 200 kg up the slope
To find the theoretical force required to push the object up the slope, we can use the formula:
F = force required
mass = 200 kg
g = acceleration due to gravity = 9.81 m/s²
μ = coefficient of friction (assume 0 for a frictionless surface)
normal force = mass * g = 200 * 9.81 = 1962 N
frictional force = μ * normal force = 0
F = (m * g * sin(angle)) + frictional force
We already have the angle (18.4°) and need to convert it to radians:
angle_rad = 18.4 * (π/180) ≈ 0.321 radians
Now we can find the force required:
F = (200 * 9.81 * sin(0.321)) + 0
F ≈ 1032 N
So, the theoretical force required to push a 200 kg object up the slope is approximately 1032 Newtons.