Question

A 40-kg crate is pulled across the ice with a rope. A force of 100N is applied at an angle of 30° with the horizontal. Neglecting friction calculate the acceleration of the crate.

Answer 1

Here is our scenario.

We should first note that the box only accelerates in the x-direction. There is no vertical acceleration because it is not being lifted off of the ground. We can verify this by seeing that the weight of the box is greater than that of the vertical component of the force being applied. Mathematically,\[\huge W>F_y \implies mg>100 \sin(30^o)\]\[\huge392\text{N}>50\text{N}\]

Moving forward, we now calculate the acceleration of the box by analyzing the forces solely in the x-direction. We know from Newton's Law that\[\huge \sum F=ma \implies \sum F_x=ma_x\]
Therefore
\[\huge ma_x=F_x\]Where \(F_x\) is the horizontal component of the applied force. Now solve for \(a_x\)