# How can a vertical force cause motion at an angle?

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]

• 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.

An object of mass $m$ on an incline with friction experiences the following three forces:

1. Gravity
2. Normal force
3. Friction

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!

It's easier to understand if you ignore friction for a minute.

Suppose you're on a road on a hill which has iced over. Anything that goes on the road, slides down because there's no friction. It is gravity that is pulling the objects down, even though they're following the surface of the hill, instead of going straight down.

In fact the hill is the very thing that stops them from going straight down.

So the hill, in stopping the objects falling straight down and forcing them to slide along the surface, is also exerting a force on the objects. This force (the normal force) is at a right angle to the surface (normal is just another word for right angle).

If you then add gravity, normal force (and, when you're ready, friction), you should get the total force.

In your problem, there are 3 forces on the block: force of gravity, friction and the normal force. To find the net force, we add the three (using vector addition).

In your coordinate system, The force of gravity is $[0, -9.8]$ and the force of friction is $[-1.3, 0.75]$ as you calculated.

To calculate the normal force, we're going to use the fact that the block doesn't accelerate perpendicular to the incline. This tells us that the normal force (since it acts perpendicular to the incline) must exactly cancel out the other forces acting in this direction. The other forces are gravity and friction. Friction doesn't have any component perpendicular to the incline, but gravity does. The component of gravity along this direction is given by $Mg \cos 30^\circ$. Hence the normal force must be equal to this in order to exactly cancel this out. So now you know it's direction and also its magnitude. In your coordinate system, it would be given by $[4.24, 7.35]$.

In your question, you ask: Shouldn't the $y$-component be $9.8$ because it doesn't accelerate in the vertical direction? It does accelerate in the vertical direction. If you look at it's acceleration vector (which would be along the incline), you will see that this vector has a component along the negative $y$-axis. That is why the normal force is a less than $9.8$ because otherwise this component vector would be $0$ and hence it could only have accelerated in the $x$-direction. I may have not explained this part clearly so if you have any doubts, feel free to ask.