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2 years ago

How would you find force of the ping pong ball rolling down the track?

Physics

1 answer:

Alina [70]2 years ago

6 0

<h3>Force of the ping pong ball rolling down the track:</h3><h3 /><h2>Answer</h2><h3>Weight </h3>

a) weight's vertical component = Normal upward force

b) weight's horizontal component = Friction force = (mass of ball)(acceleration)

These forces depend upon the track,

1) inclined or horizontal

2) steepness.

<h2>Explanation</h2>

The force of gravity points straight down, but a ball rolling down a ramp doesn't go straight down, it follows the ramp. Therefore, only the component of the weight which points along the direction of the ball's motion can accelerate the ball.

<em>weight's horizontal component = Friction force = (mass of ball)(acceleration)</em>

The other component pushes the ball into the ramp, and the ramp pushes back.

If the ramp is horizontal, then the ball does not accelerate, as gravity pushes the ball into the ramp and not along the surface of the ramp.

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3 years ago

Read 2 more answers

A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi

Alex_Xolod [135]

Magnitude: 12.1 N.
Direction: 17.0° to the 8 N force.

<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

$\displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .
$\displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}$ .

The sum of forces on each direction will be the resultant force on that direction:

Resultant force parallel to the 8 N force: $(8 + \dfrac{5\sqrt{2}}{2})\;\text{N}$ .
Resultant force normal to the 8 N force: $\dfrac{5\sqrt{2}}{2}\;\text{N}$ .

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

$\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N}$ (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

$\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491$ .

Find the size of the angle using inverse tangent:

$\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree$ .

In other words, the resultant force is 17.0° relative to the 8 N force.

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