# How does acceleration carry over when the tension force is at an angle?

I wanted to know how the acceleration of two blocks are affected when the tension force is at an angle. So I have this example:

Two blocks are connected by a massless rope with an angle of θ attached to the blocks' centers of gravity, on a frictionless plane. Another rope is applying a force (Fₓ) to the block with a mass of m₂. My question is, how is the net acceleration (a₁) of the block with a mass of m₁ related to the net acceleration (a₂) of block m₂? Is a₁ = a₂ₓ (just the horizontal part of a₂ₓ), or a₁ = a₂? Or is the idea that the two accelerations are equal not work in this situation?

In other tension examples I've, the tension always pulls perpendicular to whatever incline the boxes are on, so I'm confused if this situation would be much different.

• The block either accelerates or it doesn't. It only has one acceleration. There is no such thing as a net acceleration. – Johnathan Gross Sep 20 '17 at 3:36
• Have you drawn an actual free body diagram for each of the blocks (showing the specific forces acting on each), and, based on this, have you written down a force balance equation for each of the blocks? – Chet Miller Sep 20 '17 at 11:43

Are either of the blocks falling through the table or lifting off? No, so they both have a vertical acceleration of $0$.

Is the rope stretching or shrinking? No, so the blocks stay a constant distance from each other, meaning their horizontal acceleration is the same.

• Well I was thinking that if the angle was large enough, or m1 massive enough, the second block could have a vertical acceleration and move upwards. Would the horizontal acceleration still be exactly the same between the two blocks? – Rick Sep 20 '17 at 3:42
• No, the larger block would still have $0$ vertical acceleration and the smaller block would have a nonzero vertical acceleration. But then you're dealing with a completely different problem. – Johnathan Gross Sep 20 '17 at 3:50
• I get that the larger block wouldn't have any vertical acceleration. The rope is pulling it towards the floor and the normal force of the floor would prevent the block from accelerating downwards. What I just wanted to know that since the rope pulls the block upward the vertical component of the tension force could be large enough to give the smaller block vertical accelerations. Would the acceleration of the smaller block still be proportional to the larger block's acceleration. Could I say that the horizontal acceleration of the smaller block was cos(α) of the acceleration of m1? – Rick Sep 20 '17 at 4:11
• A proper handling of this situation is beyond the physics you learned so far. But you probably could answer a question like that if it was constrained in some way to the physics you know. – Johnathan Gross Sep 20 '17 at 6:37

## TL:DR

If the vertical component of tension is not large enough to overcome $F_g$ on $m_2$: $a_1 = a_2 = a_{2_x}$, otherwise read below.

how is the net acceleration ($a_1$) of the block with a mass of $m_1$ related to the net acceleration ($a_2$) of block $m_2$? Is $a_1$ = $a_{2_x})$ (just the horizontal part of $a_2$), or $a_2 = a_2$?

$$T_y = F_x * sin\theta$$ $$a_1 = a_2 \text{ if } T_y \le m_2 * g$$

If the above is not true, than the vertical component of the tension on $m_2$ will be greater than the force of gravity on $m_2$, and this will result in $m_2$ accelerating upwards.

I'm not really qualified to explain this so I will describe this as I understand it (physicists beware, I'm going to miss out on some stuff (torque and angular acceleration, I think)).

If $F_y > m_2 * g$ then, as $m_2$ accelerates upwards, $\theta$ will shrink towards $0$. While $\theta$ is shrinking towards $0$, the same $F_x$ is being applied; however, the $x$ component of $T$ is growing, and the $y$ component is shrinking.

While $T_y \lt m_2 * g$, $m_2$ will be accelerating upwards, gaining vertical speed. Once $T_y \gt m_2 * g$, $m_2$ will begin accelerating downwards (note that the point ($R$) at which $T_y = m_2 * g$ will occur before $\theta = 0$ ).

The result of this would be that as $m_2$ oscillates around $R$:

1. if $\theta$ is approaching $0$: $a_{2_x}$ (and $a_{1_x}$) are increasing.
2. if $\theta$ is receding from $0$: $a_{2_x}$ is decreasing.
• However, the velocity of $m_1$ is not required to be the same as $m_2$ as they decelerate, because, given that the masses are attached by a rope, $m_1$ can approach $m_2$ from behind. This means that in a situation where $m_1 \gt m_2$ (difference in inertia), and $F_x$ is an appropriate value, $m_1$ may approach (or even pass) $m_2$. This is roughly comparable to how a crossbow works (but the PROPER description of this situation (with angular acceleration) would probably better describe something like a trebuchet)...
• Also, if $\theta$ is receding from $0$ in the negative direction (upwards) and the circumstances are correct, $m_1$ may be pulled upwards by $m_2$; however, this would require much more force than simply causing $m-1$ to rotate under $m_2$.