Question

At one moment during a walk around the block, there are four forces exerted upon Fido - a 10.0 kg dog. The forces are:

Fapp = 67.0 N at 30.0° above the horizontal (rightward and upward)
Fnorm = 64.5 N, up
Ffrict = 27.6 N, left
Fgrav = 98 N, down

Resolve the applied force (Fapp) into horizontal and vertical components. Then add the forces up as vectors to determine the net force.

Answer 1

Step-by-step explanation:

Given that,

Fapp = 67.0 N at 30.0° above the horizontal (rightward and upward)

Fnorm = 64.5 N, up

Ffrict = 27.6 N, left

Fgrav = 98 N, down

(a) Resolution of the applied force:

Horizontal component,

`F_x=F\cos\theta\\\\=67* \cos(30)\\\\=58.02\ N`

Vertical component,

`F_y=F\sin\theta\\\\=67* \sin(30)\\\\=33.5\ N`

(b) Net horizontal force :

`F_x=58.02+(-27.6),\ \text{frictional force act in opposite direction of motion}\\\\F_x=30.42\ N`

It is positive, it will act in right side

Net vertical force :

`F_y=33.5+64.5+(-98),\ \text{gravitational force acts in downward direction}\\\\F_y=0`

Hence, it is clear that the net force is in horizontal direction i.e. 30.42 N due right side.