# How to find the acceleration of a cart sliding over a wire?

The problem is as follows:

The diagram from below shows a cart has a mass of $$2\,kg$$ is carrying a solution of sucrose as ballast which has a mass of $$6\,kg$$. Below the cart there is a brass sphere hanging by another wire as indicated in the diagram. The sphere has a mass of $$2\,kg$$. The whole system is supported by rings to a very long wire acting as a rail. The apparatus can slide over the wire by means of the rings. Assuming there is no friction between the rings and the rail, find the magnitude of the horizontal force which must be applied to the cart. The alternatives given in the book are as follows:

$$\begin{array}{ll} 1.&50\,N\\ 2.&60\,N\\ 3.&75\,N\\ 4.&85\,N\\ 5.&95\,N\\ \end{array}$$

What I did to solve this problem is summarized in the first diagram shown below. But the reason of my confusion is how exactly should I use the tension and the weight of the bob in the wire?.

The tension has a vertical an horizontal component, but how about the weight?. In my first attempt I only used the vertical component of the tension and equated this with the weight of the bob. I could spot that to get the force, all that is required is to get the acceleration of the system and this can be found by getting the centripetal acceleration which has been excerted to the bob. In this case the centripetal acceleration can be decomposed in two components, so the horizontal component can be used to get the required acceleration and getting the force by Newton's second law as follows:

$$F=m_{total}\cdot \vec{a}_{centripetal}$$

Therefore by using the vectors in the graph this leads to:

$$T\sin53^{\circ}=\vec{w}$$ $$T=\frac{mg}{\sin 53^{\circ}}$$

$$F_{c}=\frac{m_{bob}v^2}{R}$$

So here's where it comes my doubt as I'm assuming that solely the Tension is equal to the centripetal force.

$$T=F_{c}$$

Hence:

$$\frac{mg}{\sin 53^{\circ}}=\frac{m_{bob}v^2}{R}$$

Since both masses cancel this leads to:

$$\frac{g}{\sin 53^{\circ}}=\frac{v^2}{R}$$

$$F=m_{total}\cdot a_{c} \cos 53^{\circ}$$

$$F=(2+2+6)\cdot \frac{g}{\sin 53^{\circ}}\cdot \cos 53^{\circ}$$

$$F=10\cdot 10 \cdot \cot 53^{\circ}= 100\cdot \frac{3}{4}=75\,N$$

$$F= 75\,N$$

which according to my book it is the correct answer. But I don't feel very sure that's the right way to get to it. I felt that I just landed on it by trial and error.

My second attempt was to assume that the forces acting are the Tension and the weight (Refer to the second drawing in the image from above).

In this case what I assumed is that the centripetal force is affected by the tension and the weight pulling in the direction of the tension as follows:

$$F_{c}=T-mg\cos 37^{\circ}$$

But since I don't know the value of the Tension I used the vertical component to get it as:

$$T=\frac{mg}{\sin 53^{\circ}}$$

Using both equations this yields to:

$$F_{c}=\frac{mg}{\sin 53^{\circ}}-mg\cos 37^{\circ}$$

$$\frac{mv^2}{R}=\frac{5}{4}mg- \frac{4}{5}mg=\frac{9}{20}mg$$

Therefore:

$$\frac{v^2}{R}=\frac{9}{20}g$$

Then inserting this into the Newton's second law I'm getting:

$$F= (6+2+2)\cdot \frac{9}{20}\cdot 10 \cdot \frac{3}{5}= 27\,N$$

But this answer doesn't appear in the alternatives. What could it be wrong with this part?. Can someone help me here?. Why there is discrepancy?.

Both answers are wrong. The angle between the wire to the sphere and vertical is constant during motion. This is where you had your mistake (no centripetal acceleration). What it means is that the cart and the sphere both move to the right with the same acceleration $$a$$ in the horizontal direction.
Then for the sphere: $$\vec T+m_s\vec g=m\vec a$$ In the vertical direction $$T\cos 37^\circ=m_sg$$ In the horizontal direction $$T\sin 37^\circ=m_sa$$ From here you can divide the two equations and get $$a=g\tan 37^\circ$$ The force $$F$$ is the only force acting on the cart-sphere system in the horizontal direction, so $$F=(m_c+m_s)a=(m_c+m_s)g\tan37^\circ\approx 75N$$
Here's a simpler answer: Looking at the angle made by the bob, we can deduce that $$\frac ag = \tan 37^\circ \implies a = 7.5 ms^{-2}$$. This is the horizontal acceleration of the entire system. Since the total mass of the system (sucrose ballast + cart + bob) is $$10 \text { kg}$$, the net force $$F = ma = 10 \times 7.5 = \boxed{75 N}$$.
• What value did you use for $\tan 37^\circ$? – Andrei Jul 17 '20 at 4:27
• $\tan 37^\circ$ is $\bf{3/4}$, so $a=7.5m/s^2$ and $F=75N$ – Andrei Jul 17 '20 at 4:37