Andrea dangles her watch from a thin piece of string while the...

Question

Andrea dangles her watch from a thin piece of string while the jetliner she is in accelerates for takeoff, which takes about 21 s. Estimate the takeoff speed of the aircraft if the string makes an angle of 24° with respect to the vertical, Fig. 4–61.
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Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Bob observes his phone dangling from a cord while the train is accelerating out of the station lasting about 18 seconds. If the cord forms an angle of 26 degrees with respect to the vertical during this acceleration, estimate the speed of the train at the end of the 18 seconds. So that's our end goal. So ultimately, we're trying to figure out what the speed of the train is at the end of the 18 seconds is and that's gonna be our final answer. So we're also given a diagram of what's going on or a figure to help us visualize what's going on in this problem. We see Bob's hand at the top left hand corner of our diagram and we can see that we have his which is a blue rectangle with a black border here dangling from his hand on a cord that is 26 degrees from the vertical. And as we can see, we have some tension force being applied to the cord, which is denoted as vector FT. And then we also can see in the center of the phone, we have our coordinate plane here denoted by red arrows here. And we can also see to the far bottom left of our figure, we have a green arrow denoting the acceleration is going in the clockwise direction. It looks like it's going to the left. Awesome. So with that in mind, we're also given some multiple choice answers. Let's read them off to see what our final answer might be. And let's also note that they're all in units of meters per second. So A is 75 B is 86 C is 90 D is 97. So first off, let us create a more detailed graphical diagram to help us better visualize our problem. So I've gone ahead and found a really detailed skeleton sketch of this problem here. And as you can see in the top right or sorry, I should say the top left of our detailed graphical diagram here, we have a blue arrow denoting the acceleration going to the left. And we can also see from where Bob's hand is to the cord that stretches that's holding his phone. We have it 26 degrees from the vertical and we have some tension force going pointing towards where his hand is. And as we could see, we have mass times acceleration and then we have mass times gravity. So mass times gravity. So thinking of a free body diagram, so mass, so the mass of the phone plus gravity is pulling downwards direction on this phone. And we also have that the tension multiplied by cosine of 26 degrees is going in our vertical upwards direction, going towards the top of our figure here. And then look at this thinking it, think of it like a triangle. OK. So use tri like use your geo geometric geometric principles. And as you can see, we have like the tension multiplied by cosine of 26 degrees. And then on our one of our legs of our triangle here because the tension think of it to help visualize it as our triangle here. The hypotenuse of our triangle is gonna be this cord that Bob is holding. So think of Bob has in his hand, this cord and the cord is attached to his phone. So this cord that's holding the phone is gonna denote our hypotenuse of our triangle. So then one of the legs on this triangle is gonna be the tension multiplied by sin of 26 degrees. And then also on this dotted line, we have mass times acceleration because you have the mass of the phone. And it's also gonna be accelerating too since the train is moving. So now that we have an idea of what's going on here. And also when we look at our detailed diagram, we can see that the tension in the cord, which is denoted as T will create two components of force. So one in the horizontal direction which is T multiplied by sin of theta and one in the vertical direction which is T multiplied by cosine theta. So the vertical force must balance the gravitational force which is mass multiplied by gravity and the horizontal force must balance with F equals ma. So the force equals mass multiplied by acceleration. So now at this stage, we need to solve for the vertical force by recalling and using the following equation that t the tension multiplied by cosine of 26 degrees is equal to mass multiplied by gravity. And similarly let us recall and use the equation to solve for the horizontal force which is tension t multiplied by sine of 26 degrees is equal to mass multiplied by acceleration. So now we must solve for the acceleration value. And in order to find the acceleration value, we must divide the horizontal equation. I should be more specific, divide the horizontal force equation by the vertical force equation to get the following function or following equation. So tangent of 26 degrees is equal to acceleration divided by gravity. So now we need to rearrange this equation to isolate and sulfur A the acceleration. And when we do that, we will get that the acceleration A is equal to gravity multiplied by a tangent of 26 degrees. Awesome. So now we can plug in all of our known variables to solve for a. So when we do that, we will find the acceleration is equal to gravity, which gravity the numerical value for that is 9.8 m per second squared multiplied by tangent of 26 degrees. So when we plug that into our calculator, we will find that it's equal to 4.78 meters per second squared. And let's note that 4.78 is our answer when we round to two decimal places. So 4.78 is our final answer when we round two decimal places and the units are meters per second squared, which is perfect because that's the standard units for acceleration. So now we can recall and use the equation to solve for the final velocity now that we know what the value for acceleration is. So we need to recall that this equation is the final velocity which we're gonna denote as VF is equal to the initial velocity, which we're gonna denote as V I plus the acceleration multiplied by time. So with that in mind, we need to note that since the train is initially at rest in the train station need to note, let's make a little note here in blue that V I which is the initial velocity will be 0 m per second means it starts from rest. Thus, we can simplify our equation to write that V equals the acceleration multiplied by time. So once again, velocity is equal to the acceleration multiplied by the time. So now at this point, we can plug in all of our known variables to solve for V which will be our final answer. So with that in mind, we know that our acceleration, we found that value to be 4.78 m per second squared. And we need to multiply it by the time which the train is moving 18 seconds as we're told in the prom itself. So when we plug this into our calculator rounding to the nearest whole number, we will get 86 m per second. And that is our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B 86 m per second. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Struggling with Physics?

Andrea dangles her watch from a thin piece of string while the jetliner she is in accelerates for takeoff, which takes about 21 s. Estimate the takeoff speed of the aircraft if the string makes an angle of 24° with respect to the vertical, Fig. 4–61.
<IMAGE>

Watch next

Forces in 2D practice set

Struggling with Physics?

Andrea dangles her watch from a thin piece of string while the jetliner she is in accelerates for takeoff, which takes about 21 s. Estimate the takeoff speed of the aircraft if the string makes an angle of 24° with respect to the vertical, Fig. 4–61.<IMAGE>

Watch next

Forces in 2D practice set

Andrea dangles her watch from a thin piece of string while the jetliner she is in accelerates for takeoff, which takes about 21 s. Estimate the takeoff speed of the aircraft if the string makes an angle of 24° with respect to the vertical, Fig. 4–61.
<IMAGE>