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Only two forces act on an object (mass 3.00 kg), as in the drawing. Find the magnitude and direction (relative to the x axis) of the acceleration of the object.

$27.24^{\circ}$

Physics 101 Mechanics

Chapter 4

Forces and Newton’s Laws of Motion

Newton's Laws of Motion

Applying Newton's Laws

Cornell University

University of Washington

Simon Fraser University

Hope College

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in this question, we have to find exploration off the subject room with respect to the axe access. So how can we do that? We begin by the composing both forces that we call these one F one in this order one at two into each components. So F one and after in principle both have two components and we can deal with it as fathers. Let me deal with the general case. So in the general case, we have to access the Y axis and the X axis. And we have a vector that points somewhere in between these two axis Say here. So let me call the director v. So this is the vector V and the vector V makes an angle that I will call Teeter with the X axis. Then how can I calculate the components off the vector V so we can do it as follows The X component off the vector V He's equals two the size off that factor VII times the co sign off Pita and the white component is the question v times the sign off Pita And that's how we deal with the situation. But what is VX and what is a V Y VX is the X component off the vector V. So imagine that the sun is shining right here. As the sun shines, it will cast a shadow below the vector. And this shadow is what we call the X component. And it's throwing if something like this. So these is the X, and this is the same idea is also true for the white component. But now there's some shines here, so it casts a shadow right here. Then this is the white component off that factor these years of your wife house who know that the white is opposite to the angle teeter. So we used the signed completed and V X is age ascends to the angle, Tita. So we used the coastline to calculate it. Okay, so now that we go back to the question and so everything So for the Force F one, how can we do this? The composition. Let me highlight these equations. You force F one component X is equals to the force f one times the co sign off the angle between the vector F one on the X axis and these angle is these one, which is 45 degrees. So these easy cause troops 60 times they call sign off 45 degrees, which then he goes to 60 square. Tough, too divided by true them. 30 square root off, too. F one Why the white component is He goes to F one times the sign off detail, which is egos to 16 times the sign off 45 degrees, which is close to 60 times square. Root off two divided by truth again because the sign off 45 years ago. So the co sign up 4 to 5 and these is equal to 30 square root of truth. Now let me reproduce it for F too. For after we have a following the component X component off, if true, is he goes to after two times the co sign off the angle between after on the X axis, but no stats. F, too, is already over the X axis. The reform, the angle that it makes with the X axis is a cost to zero. So we have a co signer zero degrees in here and the coastline after zero is it goes to one, then the x component off after physicals to 40 new terms. Now for the white component. We have the following musicals to have two times that sign off zero degrees. This sign off the road agrees is a close to zero. So there is nothing on the Y axis, and that's it. Now let me summarize these results, and then we can go back to the question. Okay, Now what can we do? So now we use new term second load to calculate acceleration. So Newton's second law says that the resulting force let me call it F R. Is equal to the mass times Acceleration knows that both the force and acceleration are factors. So these equation is actually encoding two equations, one for the X axis and another one for the Y axis. For the X axis, we have the following. The resulting force on the X axis is it goes to the mass times acceleration in the X axis, That action. Then we have the following on the X axis. We have two forces that points to the same direction. You know, I mean F one component X points in this direction and soldiers after So both point in the same direction on the direction off the X axis. So we have 30 square. A tough true plus 40 is he goes to the mass which is three kilograms times acceleration Dex direction No, no for acceleration. Index direction is because 2 30 square root of true love's 40 divided by treat and this is approximately 27 0.4 haven't six meters per second squared. Now we pursued to do the same calculation for the wider action on the white direction. We have the following the resulting force over the white access. Is it close to the mass times acceleration on that access, then the resulting force on the Y axis is 30. Square it off Truth. There is no other force on that axis than 30 square root of two is he goes to three times exploration on the Y access. Therefore, the acceleration on the Y access is he goes to 30 square root of two divided by tree, which is a ghost of 10 times the square root of truth which is approximately 14 0.1 for true meters per second squared. Now let me summarize these results again and finally go to the final answer. Now that we have built components off the acceleration vector, we can proceed to complete what is its intensity. Or in this context, where is the resulting acceleration? And what is the direction off the resulting acceleration We begin by computing. What is the resulting acceleration Note that we have here Two components. The axe and white components. A drawing would be as follows. We have one component over the X axis. So we go to the right and then we have the white component over the Y axis. Then we go up. The result is the resulting acceleration, which is Neil's longer arrow. So let me call these eight and these it would be a X and these will be a wife. This is a rectangle triangle. Then we know about using the Gregorian theory. Um, that he squared is he goes to a X squared plus y squared that a former eight easy goes to the square root off a X squared plus a Y squared flooding the values that we have for both the X and the why we got the following the resulting acceleration Is it close to the square root off 27.476 squared plus 14.142 squared. This is approximately 30 0.9 meters per second squared. So the resulting exploration is this one? No. What is the direction off these acceleration? So to complete the direction we take a look at the following these using general for any vector now Not that I know that if we divide VX my view, right? So if we do something like this View y divided by V X, we get the sign off detail divided by V Co Sign off Tita. Then the V's are simply fired and you're left with the tangent off Peter. Therefore, the tangent off the angle is he goes to the white components divided by the X component. Then we have the following a tangent off the angle. Where did that? The acceleration is pointing. He's equals to the white component 14.1 for two, divided by the ex components 27.476 And this gives us approximately 0.515 And now we can couple in what angle is this? The angle teeter is equal to the inverse tangent off 0.515 which is approximately 27 0.3 degrees so acceleration is 30.9 meters per second squared and disappointing to the direction teeter equals to 27.3 degrees with respect to the X axis. So if I draw the acceleration in this diagram, we have the following these east acceleration pointing in between F one and after. It has 2030.9 meters per second squared and it makes an angle with the X axis off 27.3 degrees.

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