I just started learning physics 3 days ago and am having trouble understanding what I am doing wrong. Can someone please explain my error(s)? Thanks!
We have a 1kg object on a plane at a 30 degree angle from horizontal. Force of friction is 1.5N. We are asked to calculate the net force.
I assume the object is moving along a line with negative slope (toward positive x and negative y quadrant). All vectors are as they would be interpreted in $\mathbb{R}^2$ Euclidean space.
I solved to get:
Force normal = [5.66, 9.8], I am assuming that the y component is 9.8N because otherwise the object would move vertically through the plane, right? I got to this assumption because I didn't understand how a vertical force, gravity, could cause motion at a non-vertical angle, so I assumed that the normal force must be pushing the object forward, rather than gravity pulling it forward.
- Force g = [0, -9.8]
- Force friction = [-1.3, .75]
adding to get
- Force net = [4.3, -.75]
which is a 4.42N force acting at 30 degree angle.
The book I am using says the force should be 3.4N. The book rotates the the x,y axes by 30 degrees, I don't really like their way. I feel my main problem is not properly understanding the relationship between the normal force and gravity, and how it leads to motion at an angle.
Best Answer
An object of mass $m$ on an incline with friction experiences the following three forces:
The normal force points away from the incline's surface and is perpendicular to it. The force of friction is parallel to the incline and points in the direction opposing the motion of the object (in this case that means it points up the incline). The gravitational force points down (in the direction of the negative $y$-axis).
Since the object is not accelerating in the direction perpendicular to the incline (otherwise it would be falling through the surface or losing contact with the surface), one concludes that the component of the gravitational force perpendicular to the incline cancels the normal force, and the net force in that direction is zero.
What remains is the component of the gravitational force pointing down the incline, and the friction force pointing up the incline. A picture and some triangle-drawing/trig should convince you that the magnitude of the component of the gravitational force pointing down the incline is
$m g \sin (30^\circ) = (1\,\mathrm{kg})(9.8\,\mathrm{m}/\mathrm{s}^2)(1/2) = 4.9\,\mathrm N$.
The friction force pointing up the incline has magnitude $1.5\,\mathrm N$. Since the component of the gravitational force along the incline points down the incline, and the friction force points up the incline, the magnitude of the net force is just
$4.9\,\mathrm N - 1.5\,\mathrm N = \boxed{3.4\,\mathrm N}$.
Let me know if anything here was confusing, and I can make more comments.
Cheers!