In the following picture, the striped square represents a box that is standing on a surface that is aligned at $\alpha$ degree angle. Box stands on top of point $T$ (that I forgot to mark on paper), $\vec{TB}$ is weight of the box (in line with gravitational acceleration) while $\vec{TC}$ is the pressure of the box on surface (90 degree to surface).

enter image description here

In the book that I read, it says:

$$\vec{TB} \times cos\alpha = \vec{TC} $$

However, I think it is wrong. Since $\angle TAC = \angle BTC = \alpha$, shouldn't it be:

$$\vec{TB} \times sec\alpha = \vec{TC}$$

My proof is:

$$\vec{TB} \times sec\alpha = \vec{TC}$$

$$\vec{TB} \times \frac{1}{cos\alpha} = \vec{TC}$$

$$\vec{TB} \times \frac{1}{\frac{\vec{TB}}{\vec{TC}}} = \vec{TC}$$

$$\vec{TB} \times \frac{\vec{TC}}{\vec{TB}} = \vec{TC}$$

$${\vec{TC}} = {\vec{TC}} $$

  • $\begingroup$ It is correct that the ratio of the segments $TB/TC=\cos \alpha$. However the sides of the triangle are not the same as the magnitudes of the forces $\endgroup$ – user126422 Feb 4 '17 at 4:25

So the gravitational force is, in the proper coordinates, $-mg~\hat y$ where $\hat y$ is a unit vector in the upward direction. Let us also say that $\hat x$ is a unit vector pointing right.

Now with constrained motions you might not want to use $\hat y$ as it neither clearly obeys the constraints (in-line with the surface) nor clearly violates them (by being perpendicular to the surface). Instead you might want to use coordinates which either point up along the ramp in the direction that motion can occur, let's call that $\hat m$, or normal to the surface, let's call that $\hat n.$ Then it does not take too much effort to see that if $\alpha$ is measured in radians that,$$\hat m = \cos\alpha~\hat x + \sin\alpha~\hat y\\\hat n = -\sin\alpha~\hat x + \cos\alpha~\hat y.$$We can also solve these for $\hat y$ remembering that the Pythagorean theorem for trigonometry says that $\sin^2\alpha + \cos^2\alpha = 1.$ So then we find that $$\hat y = \sin\alpha~\hat m + \cos\alpha~\hat n.$$Then when we want $-mg~\hat y$ in these coordinates we can simply substitute to find that its component in the normal direction is, in fact, $-mg\cos\alpha,$ just as your textbook says it is.

By the way, there is a trick that seasoned physicists use to solve these sorts of problems. Sometimes, like now, you know the magnitude of a force ($mg$) but do not know how it varies with some angle. Well, since that usually comes down to a rotation, it's generally a sine or cosine. The trick that we physicists use is that we think first about what happens when $\alpha = 0$ and then we think about increasing $\alpha$ a tiny bit.

Here you can see clearly that when $\alpha = 0$ then the whole force must be normal, and so the normal component must go like $\cos \alpha$ because $\sin \alpha = 0$ when $\alpha = 0.$ And that's also an easy way to see why $1/\cos\alpha$ must be excluded immediately: then you are saying that the normal component of any gravitational force becomes infinite as the surface gets less and less tilted, which is obviously not how gravity works (it would make your trip to the bathroom scale interesting though!).

The little change in $\alpha$ allows us to figure out whether it's $+\sin$ or $-\sin$, by saying "and when $\alpha$ starts to increase this thing starts to point in the negative direction so it must be $-\sin\alpha.$" So take a second maybe and look back at the equation above and think why the $\hat x$ component of $\hat n$ must go like $-\sin\alpha$, it is because as you increment the angle it has to start pointing to the left.


If the box is at rest on the plane, there are 3 forces acting on it, keeping it in equilibrium:

  1. The gravitational pull of earth on the box acting along your line $\vec{TB}$,

  2. the normal force of the plane on the box acting perpendicular to the plane and parallel to your line $\vec{TC}$, away from the plane, and

  3. the static friction of the plane on the box acting parallel to the plane, up the plane.

The sum of these will be zero.

It is convenient to resolve these into components which are parallel and perpendicular to the plane. Forces 2 and 3 are trivial.

Force 1, the force you asked about, must be treated as the hypotenuse of a right triangle when calculating its components. The mutually perpendicular components of this force will be, must be, less than the actual force, so your secant value, sec($\alpha$), must be wrong, by definition. The component is parallel to $\vec{TC}$, into the plane, but it is not as long as that line segment.


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