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Tatiana [17]
3 years ago
8

Three men are pulling on a tree attached with ropes to keep it from falling down. If one man is pulling the rope with a force of

370N acting at 30 degrees east of north and the second man is pulling with a force of 280N acting at 70 degrees west of south. What force should the third man pull with and in what direction? PLEASE HELP AND EXPLAIN THANK YOU SO MUCH
Physics
1 answer:
slamgirl [31]3 years ago
3 0
The correct answer to this question is <span>48.8N</span><span> on a bearing of </span><span>134.2 degrees.

In order to solve this, first we need to find the resultant force of the men pulling in the north and south directions:

F = 40 - 6 = 34N [due south - 180]

Then, find the resultant force.

R^2 =34^2 + 35^2 
R^2 = 2381
R = sqrt(2381)
R = 44.8N

The angle from the vertical is given by:
tan(x) = 35/34
tan(x) = 1.0294
angle = 45.8 degrees

Taking N as zero degrees, then it bears 134.2 degrees.


</span>
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Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

\tau=Fd=mg\cdot R

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

\tau=I\alpha

therefore, equating this to the above equation gives

mg\cdot R=I\alpha

solving for alpha gives

\alpha=\frac{mgR}{I}

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}\boxed{\alpha=1.47s^{-2}}

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

\theta=\frac{1}{2}\alpha t^2

Therefore, ar t = 4.2 s, the above gives

\theta=\frac{1}{2}(1.47)(4.2)^2

\boxed{\theta=12.97}

So how many revolutions is this?

To find out we just divide by 2 pi:

\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}\boxed{\#\text{rev}=2.06}

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

\omega^2=w^2_0+2\alpha(\Delta\theta)_{}

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

\omega^2=0+2(1.47)(12.97)w^2=38.12\boxed{\omega=6.17.}

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

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