(II) An object is hanging by a string from your rearview mirro...

Question

(II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m/s in 5.0 s, what angle θ does the string make with the vertical? See Fig. 4–46.
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Accepted Answer

Welcome back. Everyone in this problem. A small ball hangs from the ceiling of a bus by a cord as the bus gradually speeds up from a standstill to 24 m per second in four seconds. What angle is formed by the cord with the Vertica? For our answer choices? A says the angle is one degree B 31 degrees C 45 degrees and D 59 degrees. No. First let's try to figure out what's happening with our uh card in our boss. OK. So here we have our card right here. And for our card, guess what? Let's consider the free body diagram of the card and the hanging bar. OK. So if I were to reproduce this diagram above uh the drawing here, OK. So we have our card and we have our hanging ball. OK? And for our ball, we know that it makes an angle phi with the vertical and the card. Now, if we think about the forces that are acting here, the tension of the card must be acting in this direction. So that would be T OK? And since our tension is acting in that direction, OK, then that means it's the components of the tension t would be acting westward here, TX and North TUI. Now, if we think about the ball as well, we know that it has its own weight acting downwards vertically and that weight would be MG. OK. I think I can make the sneer by putting Phi inside the anger that would probably be better. Now, what do we know about the components of our attention? Well, here notice that the vertical is going to form a right angle with the horizontal component of the tension and thus the horizontal component, vertical component and tension form a right triangle. Since that's the case and our vertical component is let me highlight it here. It's adjacent to our angle, Phi, then our vertical component is really the tension multiplied by the cosine of theta. Likewise our horizontal component opposite to our angle. Phi is going to be the tension multiplied by the sign of theta or the sorry, not theta phi, you know, sometimes we always say theta but really talking in this problem, we're talking about Phi. So I'm going to have to make sure I remember Phi. OK. 555. No, if we let the bus accelerate at a since the ball would be accelerating with the bus horizontally and remain stationary in the vertical direction, that means its vertical forces must sum up to zero. OK. So again, OK, the sum of its vertical forces must be zero. And if we look at our diagram for its vertical forces. T class five is going upward positive while its weight MG is going downwards negative. So T classify minus MG equals zero. That means T classify equals mg. Now again, we don't know our tension and we don't know the mass of the bar. So we still need something else to figure out our anger fight. But this is pretty important. This is an important result. So let's keep it in mind for no. What else do we know? Well, when we're talking about forces, if we recall Newton's second law force is directly proportional to acceleration. And in this case, we have the acceleration of the bus. And recall that since force is directly proportional to acceleration by the formula force equals mass multiplied by acceleration, then in this case or force in the horizontal direction which is T sine Phi is going to be equal to the mass multiplied by the horizontal acceleration. No, if we compare, let's call this formula formula two. And if we compare it to the first formula we had, let's call that formula formula one, both of them have T and Phi. So if we, we could relate them by defining equations. OK. So dividing equations two by equation one. OK, then we're going to end up with T sine phi. OK. Divided by T class Phi being equal to M divided by MG. Now here on the left T cancel T the sine of Phi divided by the cosine of Phi equals the tangent of Phi. And on the right hand side, M divided by M cancels M which leaves us with a divided by G. Now, if the tangent of Phi equals a divided by G, that means Phi is going to be equal to the inverse tangent of A divided by G. Now we're almost there to figure out P we know what G is the acceleration due to gravity, but we don't know the horizontal acceleration of the bus. So again, how can we figure that out? What do we know? Well, let's look back on the information that we are given. Recall that it says the bus speeds up from a standstill to 24 m per second in four seconds. So we know the initial velocity from standstill is 0 m per second. It goes to 24 m per second and we know it does it in four seconds. Remember that acceleration is the change in velocity over time. So since we have that change in velocity where that change in velocity is 24 m per second minus 0 m per second, divided by the time four seconds, then our acceleration must be 24 divided by 4 24 m per second, divided by four seconds. Sorry, which is 6 m per second squared. So now we have the horizontal acceleration a for our bo what does that mean? Well, let me come over here and put it in black because no oops therefore, OK. Using our formula for Phi, that means Phi must be equal to the inverse tangent of 6 m per second squared, divided by 9.81 m per second squared acceleration due to gravity. I know when we put that into our calculators. OK? We get Phi to be equal to 31.45 degrees or to two significant figures, 31 degrees. So that means that means then that the angle, the angle formed by the cord with the vertical is 31 degrees. Thanks a lot for watching everyone. I hope this video helped.

Struggling with Physics?

(II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m/s in 5.0 s, what angle θ does the string make with the vertical? See Fig. 4–46.
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Struggling with Physics?

(II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m/s in 5.0 s, what angle θ does the string make with the vertical? See Fig. 4–46.<IMAGE>

Watch next

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(II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m/s in 5.0 s, what angle θ does the string make with the vertical? See Fig. 4–46.
<IMAGE>