"A 23kg kg child goes down a straight slide inclined 38∘ above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30N force as shown in the figure.(Figure 1)

How large is the normal force of the slide on the child?"

Here is the solution for the given problem above:

Normal force = component of child's weight that acts normal to slide - 30 sin 38°
Normal force = mg(cos 38°) - 30(sin 38°)
Normal force = (23)(9.8)(cos 38) - 30(sin 38)
= 178 - 18.5
= 159.5 N

159.5 N is the answer.

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MrRoyal MrRoyal · Answer 2 · 2021-10-24T16:53:25+02:00

Force is simply the pull or push that acts on an object.

The normal force acting on the child is 159.12 N

The forces acting on the child are

The 30 N force at 38 degrees
The weight of the child

The weight of the child along the vertical component is:

[tex]\mathbf{Weight = mg \times cos(theta)}[/tex]

So, we have:

[tex]\mathbf{Weight = 23 \times 9.8 \times cos(38)}[/tex]

[tex]\mathbf{Weight = 177.62}[/tex]

The horizontal component of the 30 N force is:

[tex]\mathbf{F_2 = 30 \times sin (38)}[/tex]

So, we have:

[tex]\mathbf{F_2 = 18.50}[/tex]

The normal force (F) on the child is:

[tex]\mathbf{F = Weight - F_2}[/tex]

[tex]\mathbf{F = 177.62N - 18.50N}[/tex]

[tex]\mathbf{F = 159.12N}[/tex]

Hence, the normal force is 159.12 N

"A 23kg kg child goes down a straight slide inclined 38∘ above horizontal. The child is acted on by his weight, the normal force from the slide, kinetic friction, and a horizontal rope exerting a 30N force as shown in the figure.(Figure 1) How large is the normal force of the slide on the child?"

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