A uniform sphere with mass 60.0 kg is held with its center at the origin, and a second uniform s

A uniform sphere with mass 60.0 kg is held with its center...

Key Concepts

Point Mass Approximation

When dealing with uniform spheres or other symmetrically distributed masses, it is often convenient to treat them as point masses. This simplification assumes that the entire mass of the body can be considered as concentrated at its center, thereby making it easier to calculate gravitational forces using the inverse-square law without accounting for internal distribution complexities.

Gravitational Equilibrium

Gravitational equilibrium occurs when the net gravitational force on an object is zero, meaning that the individual forces cancel each other out. This concept is important in scenarios where multiple bodies exert gravitational influences on a test mass, and one seeks locations, other than at infinity, where these influences balance perfectly.

Newton's Law of Universal Gravitation

This law states that every pair of masses in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. It is the foundational principle for understanding how gravitational forces between objects operate and is mathematically expressed by the equation F = G*(m1*m2)/r^2, where G is the gravitational constant.

Vector Addition of Forces

Because forces are vector quantities, they have both magnitude and direction. When multiple gravitational forces act on an object, their net effect is found by vector addition, which involves breaking each force into its components and summing these components appropriately. This method is crucial for determining the overall gravitational force acting on a mass placed in a multi-body system.

Transcript

00:01 So this problem describes the following situation where we have one mass of the origin m1, another mass m2, three meters higher up, and then mass m3, four meters along the x axis away.

00:13 And we want to find what the gravitational force on m3 is from these other masses, and we're given values for these.

00:20 So to start, i'll fill in the rest of this as a triangle.

00:24 So if one side is three, the other side's four, and this is a right angle, we know then that the high ottenous is going to be 5 meters.

00:33 And this angle theta right here, we could calculate as 53 .1 degrees.

00:43 So i just did inverse tangent of 4 over 3 to get that.

00:48 So we want to find the force in m3.

00:51 So we'll do this in x and y components, and then we'll combine them to get the net force, and we can also find the angle.

00:59 So we'll start with the x.

01:01 X force, so it's going to be the gravitational force times m3, and then we'll have forces from 1 and forces from m2.

01:11 So for m1, that one's easy because it's right along the x -axis, so that's just going to be m1, and the distance is 4 meters, so 4 squared.

01:23 And then also in that same direction, so this is pulling to the left, this one is also going to be pulling it to the left in the x direction.

01:33 Plus m2 and its distance is 5 meters so 5 squared but then we only want to take the x component of that so we want to do sign of theta to get to opposite this distance over hypotenuse so then times sign of 53 .1 degrees and so we plug in m1 is 50 m2 is 80 and we should get then that this force is 1 .89 times 10 to the minus 10th newton's.

02:10 And that is in the negative x direction.

02:14 Those are both pulling to the left.

02:18 So for the y force now, we do the same thing.

02:21 We have g, m3.

02:23 And then mass 1 is going to have no y force since it's parallel with m3.

02:27 So we only have to worry about m2.

02:30 So we have m2.

02:32 Its total distance is 5 and then square that again.

02:35 And then rather than doing sine, now we want to do cosine to get adjacent this component of the force.

02:43 So then cosine of 53 .1.

02:50 And we should get then that is 6 .41 times 10 to the minus 11th newtons.

02:59 So we can draw our forces out here now.

03:03 So we have this y component and we have this y component.

03:08 And we have this x component and we want to find the net result of these two so f net and so we can just do that with the pythagorean theorem so we'll say f net is the square roots of fx squared plus f y squared and so when we evaluate that we should get 2 .00 times 10 to minus tenth newtons so that's the magnitude of the force we also want to give the direction so we could find this angle here on the force, and that would just be inverse tangent of fy over fx.

03:56 And when we do that, we should get 18 .73 degrees.

04:03 But to give kind of the more standard answer, usually we want to give the angle above the x -axis.

04:08 So this is kind of pointing like this.

04:12 And instead we want to give this angle here.

04:16 So that'll just be 180 minus the angle we found.

04:18 So 180 minus 18 .73 tells us that this is about 161 degrees above the x -axis...

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