Question

Two forces act on a 16-kg object. The first force has a magnitude of 68 N and is directed 24° north of east. The second force is 32 N, 48° north of west. What is the acceleration of the object resulting from the application of these two forces to the object?

Answer 1

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

$R= \sqrt{(R_{x})^{2}+(R_{y})^{2} }$

$R= \sqrt{(40.71)^{2}+(51.43)^{2} }$

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:

$\alpha = tan^{-1} (\frac{R_{y} }{R_{x} } )$

$\alpha = tan^{-1} (\frac{51.43 }{40.71 } )$

α = 51.63° north of east