7.3 ". CP A 120-kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces

step 1 We have a package, 120 kilograms, that's been pushed over two meters. And the first part asked u

7.3 ". CP A 120-kg mail bag hangs by a vertical rope 3.5 m...

Key Concepts

Energy Considerations in Mechanical Systems

Energy methods involve analyzing the change in the form of energy—in this context, the gravitational potential energy gained by an object when raised to a higher position. Recognizing energy changes helps in understanding how forces, even if not doing work directly along their own line of action, contribute to the change in the system's energy.

Work and the Dot Product

Work is defined as the dot product of force and displacement, meaning it depends on both the magnitude of the force and the component of the displacement in the direction of the force. In situations where the force is perpendicular to the displacement, no work is done, whereas forces with a parallel component to the displacement perform work.

Force Equilibrium

This concept involves understanding how forces balance in a static situation. When an object is held in place by multiple forces, such as gravity and an externally applied force, the net force on the object is zero. Analyzing equilibrium requires constructing a free?body diagram and setting the sum of the forces in each direction equal to zero.

Vector Decomposition

When forces act at angles, they are commonly broken into orthogonal components (typically horizontal and vertical). This decomposition allows the analysis of how each component contributes to the overall balance of forces, such as finding the horizontal force needed to counteract the displacement while the vertical component balances the weight.

Transcript

00:01 We have a package, 120 kilograms, that's been pushed over two meters.

00:05 And the first part asked us to figure out what is the force needed.

00:11 What is the force applied to this package to keep it at its current location? what is this force? first thing to do is to draw a nice, neat force diagram of the package.

00:24 So i'm going to draw dot.

00:25 This represents the, actually let's do this down here.

00:28 Draw dot.

00:29 This represents this package.

00:31 And let's label all the forces that are acting on it.

00:33 We've got the force due to gravity on this 120 kilogram package, forced due to gravity.

00:42 And we can actually calculate it, since it's 120 kilograms, it's going to be mass times the acceleration.

00:48 So 120 kg times 9 .8 meters per cent.

01:00 2nd squared.

01:02 All right.

01:03 We also have on this package, we've got a tension force from this rope that's pointed up at this angle.

01:10 So we can draw this tension force upwards at an angle, approximately similar.

01:15 I mean, just call that tension.

01:18 And we've got this force of the push, force of the push from the mail worker.

01:25 I'm just going to call that f.

01:28 Okay, we want to find f.

01:30 Well, we know that the forces in this, since this is a stationary optioned push over, and now it's just being held there, what is the magnitude of this? we know all the forces in the y direction, all forces in the x direction have to equal each other.

01:43 But before we get there, one thing that we have to solve for is we need to figure out what is this angle theta, so we can carry it over to our diagram over here.

01:56 Because that angle there is going to be the same as this angle here, which will be necessary to solve it.

02:01 A 3 .5 meter long rope and a 2 meter long displacement or 2 meter displacement.

02:11 So if we kind of connect these together and imagine drawing a right triangle here, we can find theta using the inverse sine function.

02:20 So theta is going to be equal to the inverse sign, inverse sign of the opposite side, which in this case, would be our two meters divided by our 3 .5 meter hypotenuse.

02:41 And that gives us an angle of 34 .8 degrees.

02:51 Okay.

02:52 So this theta, these two thetas, are the same.

02:56 Well, we know that the, let's see, the force, that this tension force in the, the y direction has to be equal to the force through the gravity.

03:08 So i'll even write that tension in the y has to be equal to our forced gravity down below since all the y forces have to balance out.

03:19 So this component in the y direction has equal to force through gravity.

03:24 And that the tension in the x direction, this component here, this tension in the x direction is going to be equal to our total four or our force from the worker.

03:38 These two forces in the x direction have to bounce each other out.

03:43 Well, we can relate tension x and tension y to each other...

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