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

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.

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Lệ Thu 6 months 2021-07-17T08:02:56+00:00 1 Answers 73 views 0

Answers ( )

    0
    2021-07-17T08:04:40+00:00

    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\times \cos(30)\\\\=58.02\ N

    Vertical component,

    F_y=F\sin\theta\\\\=67\times \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.

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