1. It has been suggested that rotating cylinders about 11 mi long and 5.2 mi in diameter be placed in space and used as colonies. What angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth?

2. If the torque required to loosen a nut that is holding a flat tire in place on a car has a magnitude of 52.0 N·m, what minimum force must be exerted by the mechanic at the end of a 37.0 cm lug wrench to accomplish the task?

3. A coordinate system (in meters) is constructed on the surface of a pool table, and three objects are placed on the table as follows: a 9.0 kg object at the origin of the coordinate system, a 10.0 kg object at (0, 2.0), and a 17.0 kg object at (4.0, 0). Find the resultant gravitational force exerted by the other two objects on the object at the origin.

magnitude:
direction:

We will be happy to critique your work after you have shown some effort.

Hints:
1. the length does not matter
2. Torque = force * (lever arm)
3. Compute the forces between each the origin object and each of the other two, and add them vectorially. Use Newton's universal law of gravitation

1. To find the angular speed required for the rotating cylinder, we need to set the centripetal acceleration at its surface equal to the free-fall acceleration on Earth.

Let's start by finding the centripetal acceleration at the surface of the cylinder. The centripetal acceleration is given by the formula:

Ac = ω^2 * r

Where Ac is the centripetal acceleration, ω is the angular speed, and r is the radius of the cylinder (which is half the diameter).

The free-fall acceleration on Earth, denoted by g, is approximately 9.8 m/s^2.

To set the two accelerations equal to each other, we have:

Ac = g

ω^2 * r = g

Substituting the given dimensions of the cylinder (length = 11 mi, diameter = 5.2 mi), we convert them to meters:

Length = 11 mi * 1609.34 m/mi = 17703.74 m
Diameter = 5.2 mi * 1609.34 m/mi = 8366.068 m

Radius, r = diameter/2 = 8366.068 m/2 = 4183.034 m

Now, we can solve for the angular speed (ω):

ω^2 * 4183.034 m = 9.8 m/s^2

ω^2 = 9.8 m/s^2 / 4183.034 m

ω = √(9.8 m/s^2 / 4183.034 m)

ω ≈ 0.021 rad/s

Therefore, the angular speed required for the rotating cylinder to have a centripetal acceleration equal to the free-fall acceleration on Earth is approximately 0.021 rad/s.

2. To find the minimum force required to loosen a nut using a lug wrench, we can use the equation:

Torque = Force * Lever arm

Given the torque value of 52.0 N·m and the length of the lug wrench arm as 37.0 cm (or 0.37 m), we can rearrange the equation to solve for the force:

Force = Torque / Lever arm

Force = 52.0 N·m / 0.37 m

Force ≈ 140.54 N

Therefore, the minimum force that must be exerted by the mechanic at the end of the 37.0 cm lug wrench to loosen the nut is approximately 140.54 N.

3. To find the resultant gravitational force exerted by the other two objects on the object at the origin, we can use the formula for the gravitational force between two objects.

The formula for the gravitational force (F) between two objects is given by:

F = G * (m1 * m2) / r^2

Where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

Given that G is a constant (G ≈ 6.67 × 10^-11 N(m/kg)^2), we need to calculate the gravitational forces exerted by the two other objects individually and then combine them vectorially to find the resultant force.

For the first object at the origin (mass = 9.0 kg), the distance (r) between the objects is 0 since they are at the same location. Therefore, the resultant force exerted by this object is 0.

For the second object (mass = 10.0 kg) located at (0, 2.0) on the coordinate system, the distance (r) can be calculated using the Pythagorean theorem:

r = √(x^2 + y^2)

Substituting the values:

r = √(0^2 + 2.0^2)
r = √(4.0)
r = 2.0 m

Now we can calculate the gravitational force exerted by the second object on the object at the origin:

F2 = G * (m1 * m2) / r^2
F2 = (6.67 × 10^-11 N(m/kg)^2) * (9.0 kg * 10.0 kg) / (2.0 m)^2

Similarly, for the third object (mass = 17.0 kg) located at (4.0, 0) on the coordinate system, the distance (r) is 4.0 m.

F3 = G * (m1 * m2) / r^2
F3 = (6.67 × 10^-11 N(m/kg)^2) * (9.0 kg * 17.0 kg) / (4.0 m)^2

To find the resultant force, we need to add the vectors F2 and F3 using vector addition. We can break down the forces into their x and y components and sum them separately:

Fx = F2x + F3x
Fy = F2y + F3y

Finally, we can calculate the magnitude and direction of the resultant force using the sum of the squares of the x and y components:

magnitude = √(Fx^2 + Fy^2)
direction = arctan(Fy / Fx)

You would need to plug in the numerical values for F2x, F2y, F3x, and F3y calculated from the above equation and then solve for the magnitude and direction of the resultant force.