The forces in FIGURE EX6.9 act on a 2.0 kg object. What are the values of ax and ay, the x- and y-components of the object's acceleration?

Question

Diagram showing forces acting on a 2.0 kg object on an incline with angles and force values labeled.

Accepted Answer

Hey, everyone in this problem, we have a 1.25 kg box being moved up an incline. The forces applied during the motion are shown in the figure below. And we are asked to calculate the horizontal and vertical components of the box's acceleration vector. Now this diagram we're given or the figure we're given has this box moving on an incline that makes an angle of 30 degrees with the horizontal. We have a 10.7 Newton force acting straight upwards perpendicular to the incline. We have a 10 Newton force acting to the right parallel to the incline, an 18 Newton force acting to the left parallel to the incline. And we have a 10.3 Newton force acting straight downwards or vertically. OK. That is not on that tilted axis. OK. So we have this tilted axis, we're taking the positive X axis to be to the left, the positive Y direction to be upwards. And that tilted axis is so that our horizontal forces are parallel to the incline and our vertical forces are perpendicular. Now, we're given four answer choices. ABC and D and they each contain different combination of vertical and horizontal accelerations. OK. The horizontal acceleration, we have the option of 1.5 m per second squared or 6.4 m per second squared. And for the vertical component, we have the choice of zero m per second squared or 1.3 m per second squared. OK? So we have each answer choice, just a combination of those options. The first thing we're gonna do is draw a free body diagram here. OK. We're looking for the acceleration and we know that we can relate the acceleration to the forces through Newton's Second law. So acting in our positive y direction, we have a 10.7 Newton force that's on our tilted axis, bye, acting in the negative horizontal direction. Again, that tilted axis 10 Newton force acting to the left on our tilted axis in Newton force and then acting straight down. We have a 12.3 Newton force and, and what we wanna do is we wanna break that 12.3 Newton force up into the horizontal and vertical components. And so we can do that, OK? We're gonna have some vertical component that's pointing downwards with our tilted axis. And then the horizontal component is going to point to the right and the angle between the force we're given and the tilted vertical axis is going to be 30 degrees. So recall, Newton second tells us that the sum of the forces and to start, we'll look in the X direction is equal to the mass multiplied by the acceleration and the extra action. So let's look at the some of the forces in the abstraction in the positive X direction. We have an 18 Newton force in the negative direction. We have a 10 Newton force. And we also have this X component of our 12.3 Newton force. Now that's on the opposite side of our angle. So that's gonna be related through the sign. So that's gonna be sine of 30 degrees multiplied by 12.3 nodes. So we get that 18 newtons minus 10 nos minus sine of 30 degrees times 12.3 newtons is equal to the mass 1.25 kg multiplied by the acceleration in the X direction. And that horizontal component, if we simplify on the left hand side, we get 1.85 mutants. On the right hand side, we have 1.25 kg multiplied by the acceleration A X dividing by 1.25 kg. We get that the acceleration A X is equal to 1.48. And the unit here we had newtons which we call is a equivalent to kilogram meter per second squared. We divided that by kilogram. So that unit of kilogram cancels out and we're left with just meters per second squared, which is exactly what we want for acceleration. So that's great. We found our horizontal acceleration 1.48 m per second squared, the answer choices are rounded to two significant digits. So if we round this to two significant digits, we get 1.5 m per second squared. This means we can eliminate answer choice C and D, OK. They have horizontal accelerations of 6.4 m per second squared. And so that is not going to be the correct answer. So we're looking at option A or B and let's look at the vertical component of the acceleration in order to figure out which one is correct. So just like for the horizontal direction, we have that the sum of the forces in the Y direction is equal to the mass multiplied by the acceleration in the Y direction. Newton. Second law, let's take a look at our free body diagram in the positive Y direction. We have a 10.7 Newton force in the negative Y direction. We have the vertical component of our 12.3 Newton force. And that is gonna be related to the cosine of that angle because it's the adjacent side. So we have 10.7 newtons minus cosine of 30 degrees multiplied by 12.3 newtons is equal to the mass of 1.25 kg multiplied by the acceleration A Y. So isolating for the acceleration, we get that A Y is equal to 10.7 newtons minus cosine of 30 degrees multiplied by 12. newtons. And all of this is gonna be divided by that mass of 1.25 kg. Ok. Again, we have newtons divided by kilograms which leaves us with meters per second squared. And we get that, that vertical component of the acceleration is going to be 0.3831 meters per second squared. Ok. So now we have both the vertical and horizontal components. Again, we're gonna round to two significant digits or one decimal place. In this case, if we do that with our vertical acceleration, it's just going to be equal to zero and that's going round to zero m per second squared. So when we look back at our answer choices, we found that our horizontal acceleration was equal to approximately 1.5 m per second squared. And the vertical component of the acceleration is approximately zero m per second squared, which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

The forces in FIGURE EX6.9 act on a 2.0 kg object. What are the values of ax and ay, the x- and y-components of the object's acceleration?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Vector Resolution

Inclined Plane Dynamics