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3 years ago

I'm completely stumped as to how to do this.

Physics

1 answer:

worty [1.4K]3 years ago

6 0

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force $\vec{\textbf{R}}$ and F1 are completely along the x-axis, we can conclude that

$F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}$

We can now solve for the magnitude of $F_3:$

$F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}$

$\:\:\:\:=529.7\:\text{lb}$

The angle $\theta$ is then

$\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}$

$\theta = 49.2°$

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You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of

Nezavi [6.7K]

Hi there!

For an object on an incline with friction being pulled, the following forces are present.

Force due to Gravity
Force due to Friction
Force due to tension

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
$\Sigma F = F_T + F_f - F_g$

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
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Now, we can express each force as an equation.

Force due to tension:

Must be solved for.

Force due to gravity:

On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)

This is expressed as:
$F_g = Mgsin\theta$

Force due to friction:

Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:

$N = Mgcos\theta$

$F_f = \mu Mgcos\theta$

Now, plug these expressions into the above equation.

$0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T$

Mg = 245 N (weight). Plug in all values:
$245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}$

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

$N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}$

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2 years ago

The breaking car had 10,000 J of kinetic energy before breaking after breaking it had 2000 J of kinetic energy. How much thermal

WINSTONCH [101]

Answer:

8000J

Explanation:

The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.

Kinetic energy loss by car = thermal energy gained by brakes.

∆K.E = ∆T.E ....1

The Kinetic energy loss by car can be expressed as;

∆K.E = K.E1 - K.E2

Initial K.E = K.E1 = 10000J

Final K.E = K.E2 = 2000J

∆K.E= 10000J - 2000J = 8000J

From equation 1,

∆K.E = ∆T.E

∆T.E = 8,000J

thermal energy gain by brakes = 8,000J

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3 years ago

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Answer:

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3 years ago

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