Answer:
a= 4.01 m/s² : magnitude of object acceleration
α = 51.63° north of east : direction of object acceleration
Explanation:
Forces acting on the object
F₁= 68 N , directed 24° north of east
F₂=32 N , directed 48° north of east
We take the axes equivalent to the given orientation of the forces in the x-y plane like this:
East : x+
west : x-
North: y+
south: y-
x-y components of the forces acting on the object
F₁x= 68 N* cos 24°= 62.12N
F₁y= 68 N* sin 24°= 27.658N
F₂x= -32 N *cos48° = - 21.41 N
F₂y= 32 N *sin48° = 23.78 N
Resulting of forces in x
Rx=F₁x+F₂x= 62.12N- 21.41 N = 40.71 N
Resulting of forces in y
Ry=F₁y+F₂y= 27.658N+ 23.78 N = 51.43 N
Magnitude of the resulting force acting on the object
[tex]R= \sqrt{(R_{x})^{2}+(R_{y})^{2} }[/tex]
[tex]R= \sqrt{(40.71)^{2}+(51.43)^{2} }[/tex]
R= 65.59 N
Calculation of the acceleration of the object
We apply Newton's second law:
F = m*a Formula ( 1)
F : Resulting force acting on the object (N)
m : mass of the object (kg)
a : total acceleration of the object (m/s²)
We replace F= R= 65.59 N and m= 16 kg in the formula (1) to calculate the acceleration of the object :
F = m*a
a= F/m
a= 65.59/16
a= 4.01 m/s²
Direction of the acceleration of the object (α )
The direction of acceleration of the object is the same direction of the resulting force acting on the object:
[tex]\alpha = tan^{-1} (\frac{R_{y} }{R_{x} } )[/tex]
[tex]\alpha = tan^{-1} (\frac{51.43 }{40.71 } )[/tex]
α = 51.63° north of east
