A cord is wrapped around the rim of a solid uniform wheel 0.250 m in radius and of mass 9.20 kg. A steady horizontal pull of 40.0 N to the right is exerted on the cord, pulling it off tangentially from the wheel. The wheel is mounted on frictionless bearings on a horizontal axle through its center. (b) Find the magnitude and direction of the force that the axle exerts on the wheel.

Video transcript

Hi everyone today, we are going to determine the magnitude and also direction of the force that the axle excerpts on the cylinder. So in this particular example, we are given that the value of the diameter of the cylinder is going to be 30, 70 m. And then we're given the weight of the cylinder which is indicated by the w right there. It's going to be 39 newton At the same time. We are actually pulling on the cylinder and exerting force on it with a force of 30 Newton, which is known in this example here. And I'm just gonna indicate this on the diagram as well, which is just going to be F ball like. So, so remember that what is being asked is the force that the axle exerts on the cylinder. And if we see here these two are actually forces acted by the cylinder itself. So remember on the previous session or recall Newton's third law where we have the law of action and reaction. Where F of action is going to be the negative of the f of reaction. And in this case as we want to determine the magnitude of the force of that is being exerted by the axle itself. We have to use this formula to determine what kind of forces are actually acting or acting on the excellent itself. So we know that to calculate the magnitude, we want to take an account of all forces in the X and Y component and we can use this square root formula right here to determine the result and force. So I'm just going to indicate the resulting force here. Well, we know that this resultant force can actually be projected into F. Y. Or the force in the Y direction and also F. X. Or divorce in the X direction like so. So let's start with the access on the X axis. We know that the only force acting is F X. And F. Ball. So using this action reaction law, we know that F X is going to be minus F. Full especially as they are on different direction like so and then next we want to look into the Y axis. We know that we have F. Y and W. So in this case F Y is going to be minus W or the minus of weight. So after doing that, we can actually plot these values into this formula right here and right down F. Or the magnitude of the force is going to be squared of X squared, which is just minus F. Bowl. Or since we are squaring it, we can just write it down as F. L squared plus F Y. Which is going to be W squared like. So and then we have the values that we have listed previously after paul is going to be 30 squared and w is going to be 39 newton. And this will actually be uh 49.2 Newton like. So yep. So now that we have indicated the actual magnitude of the force which is 49.2, Newton, we can start eliminating option, see an option D. After that we can actually determine the direction of the force using the tangent formula. So we know that tangent of the angle. Let's just in this case, let's just use the angle here. Which is just going to be um from the X. Access to the F. To the actual force magnitude itself. So we know that tangent is going to be whatever is in front over whatever is on the site, so whatever is in front is gonna be F. Y. And whatever is on the side is going to be F. X. This is going to be 39/30. And then recall that to find this ah angle right here, we have to use the inverse of tangent, which in this case the inverse of tangent is going to be our stand Or our Kiss tangent of this value right here, which is just going to be turning in over 30. And this will actually give us the value of the angle of 38.4°. So now, after doing this calculation, we know that the angle here is going to be 38.4° above the horizontal angle and away which is opposite from the direction of the ball. So our answer is going to be answer a like, so that's it