# Why consider only $x$ component of accelerating block pulled by cord at an angle?

A $5 \, \text{kg}$ block is pulled along a horizontal frictionless floor by a cord that exerts a force $F=12 \, \text{N}$ at an angle $\theta = 25 \, \text{deg}$ above horizontal. What is acceleration of the block?

Why, in looking for the acceleration, do we just do $12 \cos25 = 5(a)$? Why don't we also look at the $y$ component $12 \sin25$ and then find the magnitude of $i$ and $j$ ($x$ and $y$) by doing the whole square root of $I^2$ and $J^2$?

• Draw a correct free body diagram, then work out acceleration in the x direction and acceleration in the y direction. What is your conclusion? – David White Mar 21 at 6:06
• Note that the surface is frictionless. If friction were a factor then you would have to take the drag due to friction into account, which depends on the normal force, which in turn depends on the $y$ component of the pulling force. – gandalf61 Mar 21 at 14:09

## 4 Answers

In answer to your first question: Yes. You do that.

In answer to your second question: You'll find, if you do that, that it will come out with the magnitude of $12N$, which is what you started with. This is because $sin^2(x) + cos^2(x)=1$.

As the floor is horizontal and frictionless you only want the horizontal component of the force as that is the component of the force influencing the horizontal acceleration.

Your goal is to find the acceleration. So let's look at that. Since you are told that the box moves over the surface, you then know that this acceleration is along the x direction.

I will always start like that: Make sure you know what you are looking for, and then look in the direction where it appears.

By using Newton's 2nd Law in the x direction, you'll take the force components as you did, and equal them to $ma$. Since this $a$ is exactly what you are looking for, you don't need to do any more than this. This is enough.

Only if you have more unknowns - like if you didn't know the value of a force that is acting on it in the x direction - you will have to look into other directions also to get more equations.

To add to Steeven's answer, the vertical component (here the sine 25 component) of the applied force is used up, alongside the normal force of the ground, in balancing the weight of the block. The result of this is lowering the normal from the value it would have had if the applied force had been horizontal. In the given scenario, the vertical component of the applied force isn't enough to reduce the normal reaction to zero, i.e. to lift the block from the ground by accelerating it. In other words, the vertical component would have led to acceleration in vertical direction as well. As it is said that the block moves horizontally only, it is understood that the vertical component of the applied force is used up only to balance a part of the weight of the block, and not in accelerating it.

Here's my take. It's not good enough to say that you only look at the $x$ component because that's what you are asked to find. Newton's laws of motion always apply regardless of what the question asks for.

So let's consider the $y$ component. The free body diagram would show you that you have the weight of the object, the normal force, and the $y$ component of your external force. If you decompose your external force for its $y$ component, you get

$12 \sin{25^\circ}=5.0\:N$ up to the same sig figs as the given force.

But you still have the normal force to consider. Just like Arkya Chatterjee said, your normal force will now be reduced from this 5 N $y$ component of the external force. But it won't be reduced to zero.

Therefore, after all the forces in the $y$ direction are considered, you end up with the net force along $y$ of zero. Hence, no acceleration in that direction.