# A 4-kg sphere rests on the smooth parabolic surface 8

A 4-kg sphere rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass $m_b$ of block B needed to hold it in the equilibrium position shown.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

Let us first draw a free body diagram. How we got the angles will be explained right after.

The first step is to figure out the angle between the y-axis and the normal force applied to the sphere by the surface. Looking at the image given to us, we see at x = 0.4 m, the ball makes contact with the parabolic slope. Therefore, we must find the slope at that point to figure out the angle. To do this, we will have to take the derivative of the parabolic curve.

$y\,=\,2.5\,x^2$

$\dfrac{\text{d}y}{\text{d}x}\,=\,5x$

(Substitute x=0.4 into our derivative)

$y'\,=\,5(0.4)$

$y'\,=\,2 m$

Now, we can use the inverse of tan to figure out the angle.

$\text{tan}\theta\,=\,2$

$\theta\,=\,\text{tan}^{-1}(2)\,=\,63.4^0$

Let us now write our equations of equilibrium.

$\rightarrow ^+\sum \text{F}_\text{x}\,=\,0$

$T\text{cos}\,(60^0)\,-\,N\text{sin}\,(63.4^0)\,=\,0$ (eq.1)

$+\uparrow \sum \text{F}_\text{y}\,=\,0$

$T\text{sin}\,(60^0)\,+\,N\text{cos}\,(63.4^0)\,-\,39.24\,=\,0$ (eq.2)

Show me the free body diagram

We can now solve for T and N. Isolate for T in eq.1.

$T\,=\,\dfrac{N\text{sin}\,(63.4^0)}{\text{cos}\,(60^0)}$

(simplify)

$T\,=\,1.79N$ (eq.3)

Substitute the isolated T value into eq.2.

$1.79N\text{sin}\,(60^0)\,+\,N\text{cos}\,(63.4^0)\,-\,39.24\,=\,0$

(solve for N)

$N\,=\,19.6$ N

Substitute the value of N into eq.3 to figure out T.

$T\,=\,1.79N$

$T\,=\,(1.79)(19.6)\,=\,35.1$ N

This T value tells us the tension in the rope holding the sphere in equilibrium. This tension is applied to the sphere by the large block attached at the end of the rope. Therefore, to figure out the weight of this block, we simply divide it by the force of gravity.

$W\,=\,mg$

(Where W is weight, m is mass and g is the force of gravity)

$m\,=\,\dfrac{W}{g}$

$m\,=\,\dfrac{35.1}{9.81}$

$m\,=\,3.58$ kg

Normal force = 19.6 N

mass of block B = 3.58 kg

## 8 thoughts on “A 4-kg sphere rests on the smooth parabolic surface”

• questionsolutions Post author

We have to differentiate the equation to find the slope at the point where the ball contacts with the surface. Once you find the slope, you can find the angle by taking the inverse of tan.

Think of it this way. Slope is ∆y/∆x, which is the same as opposite over adjacent, if we take the tan inverse, we get the angle. To get the slope from a curve with an equation given, we must take the derivative. Hopefully, that helps 🙂