A 100 kg chesterfield is pushed up a frictionless ramp at a constant speed by a delivery person.
If the ramp is inclined at 25.5° to the horizontal, what horizontal force must the delivery person apply to the chesterfield?
To find the horizontal force required to push the chesterfield up the ramp at a constant speed, we need to analyze the forces acting on the object.
1. Begin by drawing a free-body diagram of the chesterfield, indicating all the forces involved.
- There is the weight force acting downward, which can be represented by the equation: Fg = mg, where m is the mass of the chesterfield and g is the acceleration due to gravity (approximately 9.8 m/s²).
- There is the normal force perpendicular to the ramp, which cancels out the vertical component of the weight force. The normal force can be calculated as: Fn = mg cos(θ), where θ is the angle of inclination of the ramp.
- Finally, there is the force of the delivery person pushing the chesterfield up the ramp. This force can be represented by the equation: F = mgsin(θ), where θ is the angle of inclination of the ramp.
2. Given the angle of inclination of 25.5°, we can now calculate the horizontal force required.
- Plug in the values into the equation: F = mgsin(θ)
- From the problem, we know the mass of the chesterfield is 100 kg, so m = 100 kg.
- The acceleration due to gravity is 9.8 m/s².
- Substitute the angle of inclination θ = 25.5° into the equation.
- Calculate the horizontal force: F = (100 kg)(9.8 m/s²)sin(25.5°).
By evaluating this expression, you'll find that the horizontal force required to push the chesterfield up the ramp is approximately 410.4 N (Newtons).