Question

What is the magnitude and direction (right or left) of the

net force acting on the box?
Top:Fn=12N. Left:Ff=3N. Right:Fp=15N. Bottom:Fg=12N​

Answer 1

12n

Step-by-step explanation:

Answer 2

Answer: 12 N to the right

Step-by-step explanation:

If we calculate the net force acting on the box, we will have:

In y-component:

`Fy_(net)=F_(n)+F_(g)` (1)

Where
`F_(n)=12 N` is the Normal force, directed upwards and
`F_(g)=-12 N` is the weight of the box (gravity force), directed downwards.

`Fy_(net)=12 N-12 N` (2)

`Fy_(net)=0 N` (3) Hence the net force in the vertical component is zero

In x-component:

`Fx_(net)=F_(left)+F_(right)` (4)

Where
`F_(left)=-3 N` and
`F_(right)= 15 N`

`Fx_(net)=-3 N + 15 N` (5)

`Fx_(net)=12 N` (6) This is the net force in the horizontal component

Therefore, the total net force acting on the box is 12 N directed to the right