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3 years ago

5. A cable is attached 32.0 m from the base of a flagpole that is about to

Physics

1 answer:

soldi70 [24.7K]3 years ago

7 0

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, $F_v$ = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

$F_v$ + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈ 4513.093·d

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

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1) Magnitude of the net force: 20.8 N

2) Angle: $35.2^{\circ}$ clockwise from positive x-direction

Explanation:

First of all, we need to find the components of the resultant force along the horizontal and vertical direction.

We observe that there are two forces acting along the horizontal direction:

$F_1=33 N\\F_2 = 16 N$

We notice that they act in opposite directions, so the net force in the horizontal direction is the difference between these two forces:

$F_x =F_1-F_2=33-16=17 N$ (to the right)

Instead, there is only one force acting in the vertical direction, so the net force in the vertical direction is:

$F_y=F_3=12 N$ (downward)

Therefore, the magnitude of the net force can be found by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{17^2+12^2}=20.8 N$

Now we have to find the angle of the resultant net force. We can do it by using the following equation:

$tan \theta = \frac{F_y}{F_x}$

where

$\theta$ is the angle, measured as clockwise from the positive x-direction (because the vertical net force is downward)

$F_y = 12 N$ is the vertical net force

$F_x=17 N$ is the horizontal net force

Solving, we find:

$tan \theta=\frac{12}{17}=0.706$

And so the angle is

$\theta=tan^{-1}(0.706)=35.2^{\circ}$

In a direction down with respect to the right.

Learn more about forces:

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3 years ago

A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y

Gennadij [26K]

Answer:

Negative z-direction.

Explanation:

We need to determine the direction of the magnetic force. Since the velocity of the proton is in the positive x direction, and the magnetic field is in the positive y direction, we know by the vectorial formula $F=q(v\times B)$ (or, alternatively, with the <em>left hand rule</em>) that the magnetic force points in the positive z-direction (also taking into account that <u>the charge is positive</u>), so the electric field should be in the negative z-direction to balance it.