Answer:
f =216.5 N
Explanation:
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)
Known data
m=50 kg : mass of the box
F=250 N : Force applied to the box
θ = 30°: angle θ of the F=250N below the horizontal
μs=0.40: coefficient of static friction between the box and the surface
μk= 0.30: coefficient of kinetic friction between the box and the surface
g = 9.8 m/s²: acceleration due to gravity
Forces acting on the box
We define the x-axis in the direction parallel to the movement of the box on the floor and the y-axis in the direction perpendicular to it.
W: Weight of the box : In vertical direction downaward
N : Normal force : In vertical direction the upaward
F : Force applied to the box: In 30° below the horizontal direction .
f : Friction force: In horizontal direction
Calculated of the weight of the box
W= m*g = (50 kg)*(9.8 m/s²)= 490 N
x-y components F
Fx = Fcosθ= ( 250 N)*cos(30)° = 216.5 N
Fy =Fsinθ = ( 250 N)*sin(30)°= 125 N
Calculated of the Normal force
∑Fy = m*ay ay = 0
N-W-Fy= 0
N-490 N-125 N = 0
N =615 N
Calculated of the static friction force (fs):
fs=μs*N= 0.4*615 N
fs= 246 N
Problem development
We have to
fs= 246 N
Fx =216.5 N
Since the minimum force Fx is less than the fs, means that the box remains at rest and we apply Newton's first law to calculate f :
∑Fx = 0
f-Fx=0
f = Fx
f =216.5 N
