A 15.0-kg ball is supported from the ceiling by rope A. Rope B pulls downward and to the side on

A 15.0-kg ball is supported from the ceiling by rope A. Rope...

Key Concepts

Static Equilibrium

In problems involving forces acting on a stationary object, the principle of static equilibrium states that the net force in any direction is zero. This means the sum of all horizontal components must cancel, as must the vertical components, which serves as the foundational concept for setting up equations to solve for unknown forces such as tension in ropes.

Free?Body Diagram

A free-body diagram is a schematic representation used to illustrate all the forces acting on an object. By isolating the object and drawing vectors for gravitational force, tension, and any additional applied forces, one can clearly visualize the force balance and apply equilibrium conditions.

Force Resolution

Force resolution involves breaking each force vector into its horizontal and vertical components using trigonometric relationships. This method is key to analyzing each component separately and setting up corresponding equilibrium equations, which simplifies the process of solving for unknown forces.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, are used to relate the magnitude of a force with its components based on the angle it makes with the vertical or horizontal axes. These functions are essential for correctly decomposing forces in problems where vectors are inclined at given angles.

Transcript

00:01 All right, so chapter 9, problem 67, a 15 kilogram ball is supported from the ceiling by rope a.

00:10 Rope b pulls downward into the side of the ball.

00:15 If the angle of a to the vertical is 22 degrees, and if b makes an angle of 53 degrees to the vertical, and you can see figure 9 -85 for this, then find the tensions in ropes a and b.

00:34 All right, so here we have our picture, right? so we have the ball there.

00:46 Yeah, we have the ball there being held by a and then b pulling it a little bit more.

00:52 So theta b is going to be larger than theta a for sure.

00:55 We have the exact numbers.

00:56 We don't even need to say that really specifically.

01:01 But yeah, so theta a is 22 degrees.

01:03 Theta b is 53 degrees and the mass of the ball is 15 kilograms.

01:07 Our goal is to find the tension in these.

01:09 And so usually when we're dealing with forces like tension, the easiest thing to do is newton's second law.

01:16 So let's see.

01:17 How do we do this? right.

01:20 So yeah, just use newton's second law, f equals ma, but only after you've drawn a free body diagram.

01:26 So you need to understand the forces going on in the situation first, right? then we can work on f equals m .a.

01:36 So if we look at the first picture here, then we can see that there's three big forces.

01:44 Happening right on on the ball not only do you have gravity pulling down on it but you also have a pulling a has his own there's a in the rope a there's going to be a tension upwards to the left and then in a rope b there's going to be a tension downwards into the right right so this is going to all go on our feed body diagram which you see here not exactly perfectly straight but i tried so just looking at just focusing on this we can see that, you know, the tensions from a and b are going to be split, and they're going to be, they can be componentized.

02:27 So they have x and y components, in other words.

02:31 And in that, we can say that the x ones basically cancel each other.

02:36 Well, they don't cancel each other, but they are going in opposite directions.

02:40 So when we sum the forces in the x direction, it's equal to zero, because the ball is supposedly not moving.

02:48 And then it's going to be equal to minus f .a .x component.

02:54 As you can see in the picture, it's pointing to the left for fa.

02:57 And then for fb, it's pointing to the right.

02:59 So minus faax plus fbx.

03:02 And so we can just rearrange that to have them equal each other.

03:05 And then the x component is if i don't, yeah, no, i have the triangle here.

03:11 So just following the triangle, if you look at, for example, theta a, what is the only that we can use.

03:20 Well, we only want the x component.

03:22 So it's pointing directly, like, directly, excuse me, pointing directly away from the angle.

03:29 So that's going to be an opposite.

03:30 And then we also have the hypotenuse.

03:32 So what's the only trick function that deals with opposite and hypotenuse? sign, right? so we want to use sign here.

03:40 So then we have, we can change the x components instead of saying f -a -x and f -b -x, we can make them sign with their corresponding angles.

03:48 So we have, instead of f -a -x, we have, instead of f -a -x, we f .a.

03:50 Sine theta a and fb, sine theta b on each of those sides.

03:57 So let's just isolate fb by itself for whenever we figure out what fa is.

04:02 And so you essentially just divide sine theta b...

Please give Ace some feedback