Answer:
(a) F = 574.3 N
(b) F = 607.4 N
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data
β= 19° : Angle of inclination of the ramp
μk = 0 : Coefficient of kinetic friction
m =180 kg :piano mass
g = 9.8 m/s² : acceleration due to gravity
W= m*g : Piano Weight
W= 180*9.8= 1764 N
X-Y axes in the inclined plane
We define the x-axis in the direction of the inclined plane ,19° to the horizontal.
We define the y-axis and in the direction of the plane perpendicular to the inclined plane.
We calculate the weight component parallel to the displacement of the piano:
Wx= W*sin19°= 1764*sin19°= 574.3 N
Problem development
a) The man pushes the piano with a force (F) parallel to the inclined plane.
We apply formula (1):
∑F = m*a , a=0 : Because the velocity is constant
F-Wx = 0
F = Wx
F = 574.3 N
b) The man pushes the piano with a force (F) parallel to the floor.
We apply formula (1):
We define the F force component parallel to the displacement of the piano (Fx):
Fx= F*cos19°
∑F = m*a , a=0 : Because the velocity is constant
Fx-Wx = 0
F*cos 19°- 574.3 N = 0
F = (( 574.3 ) / (cos 19°) )N
F = 607.4 N
