Skip to main content
Ch 10: Dynamics of Rotational Motion

Chapter 10, Problem 10

One force acting on a machine part is F = (-5.00 N)i + (4.00 N)j. The vector from the origin to the point where the force is applied is r = (-0.450 m)i +(0.150 m)j. (a) In a sketch, show r, F, and the origin.

Verified Solution
Video duration:
3m
This video solution was recommended by our tutors as helpful for the problem above.
833
views
Was this helpful?

Video transcript

Hey everyone today, we're dealing with the problem about the Cartesian plane. So we're being told that we have A door and we're exerting a force at a point on the door with a position vector of two. I negative three J. With em being the unit of the vector. Or so. The vector is the position. So it's in meters with respect to the hinge. The exerted force has given any unit vector notation Where F is equal to 15 n. or 15. Newton, I -10. Newton. J. The hinge is located at the origin of the Cartesian plane. So we're being asked to draw the position vector R. The force vector F. And the origin. So I've already gone ahead and drawn out our plane here. I've put our origin as well. And before we get started, I'd just like to note that each of these units is two. So this will be two before 68, 10, 12, 14 and so on. So with that out of the way, let's just go ahead and take a look at this. So typically the directions of the position vector and the force are represented in the Cartesian frame. And we can say that the unit I I represents the positive X direction where a negative I would be the negative X. Direction. And similarly J represents the positive Y direction, negative J represents the negative Y direction. Let's go ahead and write that into, we have positive J. Positive I negative J and negative I. So going ahead with this with this logic, let's say we write the position vector R in blue. We can say its position if I. S. X. Is I I. Is the X axis and J. Is the Y axis. This position will be to negative three. And what does that look like? Well too is this first value here in negative three would fall somewhere between two and four. Right in the middle. So the position the position vector would fall as such from the origin. Just the position vector R. Let's go ahead and take a look at the force. The force is said to be 15. I -10J. Which is just 15 negative 10. Referred to think about this as X and Y. So 15. Well we established that this last one, this last marker here is 14, so 15 should fall just beyond that and 10 negative 10 will be 2468, 10. So it'll be somewhere around here. So let's go ahead and draw that into. And our arrow, that is our force. And there you have it were only representing the directions of the vectors. That's what's being asked here. So that'll be the graphic the position vector R. The force vector F. And the origin as represented in the Cartesian plane as shown above. I hope this helps. And I look forward to seeing you all in the next one
Related Practice
Textbook Question
Under some circumstances, a star can collapse into an extremely dense object made mostly of neutrons and called a neutron star. The density of a neutron star is roughly 10^14 times as great as that of ordinary solid matter. Suppose we represent the star as a uniform, solid, rigid sphere, both before and after the collapse. The star's initial radius was 7.0 * 10^5 km (comparable to our sun); its final radius is 16 km. If the original star rotated once in 30 days, find the angular speed of the neutron star.
2537
views
1
rank
Textbook Question
Calculate the torque (magnitude and direction) about point O due to the force F in each of the cases sketched in Fig. E10.1. In each case, both the force F and the rod lie in the plane of the page, the rod has length 4.00 m, and the force has magnitude F = 10.0 N. (a)

1590
views
Textbook Question
A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long, and he exerts a 17.0-N force at the end of the handle at 37° with the handle (Fig. E10.7).

(a) What torque does the machinist exert about the center of the nut?
1122
views
Textbook Question
Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?

1690
views
1
rank
1
comments
Textbook Question
A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.500 m. After the system is released, find

(b) the acceleration of the box, and
4970
views
Textbook Question
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.

2568
views
2
rank