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3 years ago

Calculate the tension on the rope. A. 57.89 N B. 32.73 N C. 69.84 N D. 12.55 N

Physics

1 answer:

Ratling [72]3 years ago

3 0

Hi there!

$\large\boxed{\text{B. 32.73N}}$

To calculate the tension, we must calculate the acceleration of the system.

Begin with a summation of forces:

∑F = -M₁gsinФ + T - T + M₂g

Simplify and solve for acceleration: (Tensions cancel out)

$a = \frac{-M_1gsin\theta + T - T + M_2g}{M_1+M_2}$

Plug in values. Let g = 10 m/s²

$a = \frac{-3(10)(sin30)+8(10)}{3+8} = 5.91 m/s^{2}$

Now, to find tension, let's sum up the forces acting on ONE block. For simplicity, we can look at the hanging block:

∑F = -T + W

ma = -T + W

Rearrange to solve for T:

T = W - ma

We know the acceleration, so plug in the values:

T = (8)(10) - (8)(5.91) = 32.73 N

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To do that we will use the following formula:

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Where:

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We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

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Solving the operations:

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Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

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3 years ago

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astraxan [27]

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3 years ago