# How to find the acceleration of a car ascending in an incline when a sphere makes an angle in a quarter of a circle cavity?

The problem is as follows:

The figure from below shows a car going up in an incline. The car has a circular cavity on it where there is a small sphere over it. Assume the circular surface has negligible friction. Given these conditions find the acceleration in meters per second square which the wagon must have so that the ball takes the position as shown in the diagram. The alternatives given are as follows:

$$\begin{array}{ll} 1.&9.80\,\frac{m}{s^2}\\ 2.&8.33\,\frac{m}{s^2}\\ 3.&6.25\,\frac{m}{s^2}\\ 4.&5.66\,\frac{m}{s^2}\\ 5.&4.57\,\frac{m}{s^2}\\ \end{array}$$

In this problem I'm not sure how to proceed. But my instinct tells me that the acceleration of ascention must be equal to the centripetal acceleration of the ball. But I'm confused exactly at how show I make FBD or something similar to see how forces are acting on the body, therefore a draw or sketch would be appreciated in order to spot exactly the justication of the following calculations.

If I were to ignore the thing that the wagon is on an incline, the bob would have:

$$mg\cos 37^{\circ}=\frac{mv^2}{R}$$

In this case the masses cancel, and the answer would be just $$g\cos 37^{\circ}$$. But this doesn't convince me much. Can someone help me here?.

• As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. – ole Feb 2 at 15:17
• The fact that the cavity is a circular arc just means that you can use the labeled angle on the figure to determine the direction of the normal force at the point where the sphere rests on the surface of the cavity. The problem could equally well have told you there was a glass of water in a cup holder in the car and could have described the angle of the surface the the water in the glass. – David K Feb 2 at 15:22
• @DavidK Interesting observation I totally overlooked that such fact would happen if a glass of water or any liquid would be put in the wagon. Initially I thought that there was a centripetal acceleration because a bob was over a circular surface but I think in this given context it mentions that the bob is held in that position and not moving and because of such will not be a centripetal acceleration. Am I right with this assumption?. – Chris Steinbeck Bell Feb 2 at 19:29
• That is my interpretation. We’re supposed to assume the sphere found its equilibrium position and stays there. In practice I think it would be oscillating due to the transition into the configuration shown, but that makes the problem way more complicated than it was meant to be. – David K Feb 2 at 21:13

As the car accelerates upward along the ramp with $$\vec a$$, the small sphere experience an effective constant gravity as $$-\vec a$$. Together with the downward $$\vec g$$ they make a net effective gravity $$\overrightarrow{g_{\text{eff}}}$$ that forms $$16^{\circ}$$ with the vertical line, as indicated in the diagrams below. This angle $$16^{\circ}$$ is understood as the (approximate) right triangle having the $$7$$-$$24$$-$$25$$ Pythagorean triple. Similarly, the angle $$16^{\circ}$$ is associated with the $$3$$-$$4$$-$$5$$ Pythagorean triple.

Denoting the magnitude of the car acceleration as $$|\vec a| = a$$ and the usual downward gravity as $$|\vec g| = g = 10$$ , the rightmost diagram reads

$$\frac{ \text{horizontal short leg} }{ \text{vertical long leg} } = \frac{\frac45 a}{ g + \frac35 a} = \frac7{24}$$ Solve for $$a$$ we have $$\frac{96}5 a = 7g + \frac{21}5 a \implies \frac{75}5 a = 70 \implies a = \frac{70}{15} \approx 4.6667$$

Thus the answer is the $$5$$th option, where the minor discrepancy is either a typo, or due to the approximated $$g=10$$ instead of $$9.80$$.

• Thanks for doing that nice diagram. It makes sense what you established because it links with the provided information. I totally overlooked the fact that the vector of the acceleration is moving along the incline and not paralell to the ground where the incline is supported. This was the source of my confusion. – Chris Steinbeck Bell Feb 2 at 19:26
• Regarding the answer I also believe it is a typo. Becuase your solution is consistent with what others have arrived. – Chris Steinbeck Bell Feb 2 at 19:27
• Actually if you use $g = 9.80$ then the answer is "exactly" $a = 4.57$, but of course it is still might be a typo. – Lee David Chung Lin Feb 2 at 19:30
• Thanks for noting that. I totally overlooked that fact. Out of curiosity. What software did you used to draw your FBD?. – Chris Steinbeck Bell Feb 2 at 19:39
• Here I used GeoGebra. There are several commonly used free apps that are all very similar and equally good (useful) so you can pick whichever you like. – Lee David Chung Lin Feb 2 at 19:42

I think Lee David Chung Lin's answer is probably how you were intended to solve this problem, especially given the hint to assume that $$\sin 16^\circ = \frac7{25}.$$

I prefer the following acceleration diagram, however, decomposing the accelerations into components parallel to the ramp and perpendicular to the ramp: Here we can make the usual assumption that $$\sin 37^\circ = \frac35$$. But there is no need to refer to the angle $$16^\circ$$ in any way.

To fill out the diagram, notice that we have two $$3$$-$$4$$-$$5$$ triangles, one inside the other. The only known acceleration is the hypotenuse of the smaller triangle, but you can use it to get both of the other sides of that triangle. Then you have one leg of the larger triangle and can use it to get the other sides. Finally, $$a$$ is the difference between the two collinear legs of the triangles.

• I tried to use this approach but I couldn't find a way to get to the answers. Can you add an additional hint?. I'm still stuck on this. – Chris Steinbeck Bell Feb 2 at 19:24

As far as I understand, the ball does not move along the cavity, therefore, there's no centripetal acceleration of the ball. If so, you should write down the projected forces on the horizontal and vertical axes. Forces are following: gravitation force $$mg$$, supporting force $$N$$ and D'Alambert's force $$-ma$$. D'Alambert's force $$-ma$$ is the force opposite to acting acceleration. https://en.wikipedia.org/wiki/D%27Alembert%27s_principle $$-macos37+Nsin16=0$$

$$N=\frac {macos37} {sin16}$$

$$-mg-masin37+Ncos16=0$$

$$-mg-masin37+\frac {macos37} {sin16}cos16=0$$

$$a=\frac {g} {\frac {cos37} {sin16}cos16-sin37}=\frac {gsin16} {cos37cos16-sin37sin16}=\frac {gsin16} {cos53}=\frac {gsin16} {sin37}=\frac {10*7/25} {3/5}=\frac {14} {3}=4.67$$

$$sin37=\sqrt \frac {1-cos74} {2}=\sqrt \frac {1-sin16} {2}=\sqrt \frac {1-\frac {7} {25}} {2}=\frac {3} {5}$$

Seems like there's a typo in the last alternative.

• Honestly I don't know what D'Alembert's force mean. Can you explain this part to me please?. It would help a lot if you could include a FBD to justify the equations or some sort of additional explanation of where did you obtained the equations you put there because it is not very obvious to me. Can you help me with that part please?. – Chris Steinbeck Bell Feb 2 at 19:23
• I've added the Picture and explanation. – ole Feb 3 at 15:17