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3 years ago

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A packing crate is sitting at rest on an inclined loading ramp. How does the magnitude of the force of static friction compare t

o the other forces acting on the crate?

1 answer:

eduard3 years ago

8 0

As a crate is resting on the inclined surface of the ramp we have following forces on the crate

1. Weight of the crate vertically downwards

2. Normal force due to ramp perpendicular to the surface

3. static friction force along the ramp upwards

so here we know that crate is at rest

so we can say that

normal force is balanced by component of weight of crate

so we have

$N = mg cos\theta$

also we can say that friction force is balanced by another component of weight of the crate

$F_f = mg sin\theta$

so friction force is less than the weight of the crate and it will act opposite to it

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Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

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Answer:

The acceleration is 3.16x10¹⁷ m/s².

Explanation:

First, we need to find the magnitude of the Coulombs force (F):

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Now, we can find the acceleration:

$a = \frac{F}{m} = \frac{2.88 \cdot 10^{-13} N}{9.1 \cdot 10^{-31} kg} = 3.16 \cdot 10^{17} m/s^{2}$

Therefore, the acceleration is 3.16x10¹⁷ m/s².

I hope it helps you!

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Read 2 more answers

A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.

frutty [35]

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

$\sum{F_{x}}=-Tcos\theta + F=0$ (1)

$\sum{F_{y}}=Tsin\theta - mg=0$ (2)

Where $T$ is the tension force of the rope, $m=28 kg$ the mass, $g=9.8 m/s^{2}$ the acceleration due gravity and $mg$ is the weight.

On the other hand, we can calculate $\theta$ as follows:

$cos\theta=\frac{s}{l}$

$\theta=cos^{-1}(\frac{s}{l})$

Where $s=11.1 m$ and $l=18 m$

$\theta=cos^{-1}(\frac{11.1 m}{18 m})$

$\theta=51.9\°$ (3)

Now, we firstly need to find $T$ from (2):

$T=\frac{mg}{sin\theta}$ (4)

$T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}$

$T=348.69 N$ (5)

Substituting (5) in (1):

$F=Tcos\theta$ (6)

$F=348.69 N cos(51.9\°)$

Finally:

$F=215.15 N$

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3 years ago