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.

A 4-kg sphere rests on the smooth parabolic surface

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

Solution:

Show me the final answer↓
Let us first draw a free body diagram. How we got the angles will be explained right after.

A 4-kg sphere rests on the smooth parabolic surface

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

 

Final Answers:

Normal force = 19.6 N

mass of block B = 3.58 kg

This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 3, question 3-30.

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