A 115.0 kg box is pushed by a horizontal force F at constant speed up a frictionless ramp which makes an angle of 46.0 deg with the horizontal.what is the magnitude of the normal force between the ramp and the box?
The applied force= 1167.04N
m*g = 115 * 9.8 = 1127 N. = Wt. of box.
Fn = 1127*cos46 = 782.9 N. = Normal
force.
To find the magnitude of the normal force between the ramp and the box, we can use Newton's second law of motion.
First, let's resolve the forces acting on the box along the inclined ramp:
1. The weight of the box (W): The weight is given by the formula W = m * g, where m is the mass of the box and g is the acceleration due to gravity (approximately 9.8 m/s²). So, W = 115.0 kg * 9.8 m/s².
2. The vertical component of the applied force (Fv): Since the ramp is inclined, we need to find the vertical component of the applied force. It can be calculated using the formula Fv = F * sin(θ), where θ is the angle of the ramp, given as 46.0 degrees.
Now, the normal force (N) is equal to the sum of the vertical components of the weight and the applied force:
N = W * cos(θ) + Fv
Let's calculate the values:
W = 115.0 kg * 9.8 m/s²
Fv = 1167.04 N * sin(46.0 degrees)
N = (115.0 kg * 9.8 m/s²) * cos(46.0 degrees) + (1167.04 N * sin(46.0 degrees))
By solving this equation, we can find the magnitude of the normal force between the ramp and the box, which depends on the given values of the mass of the box (115.0 kg), the angle of the ramp (46.0 degrees), and the applied force (1167.04 N).