A rope pulls a box along a floor with a force of 120 N at 25 degrees above the horizontal
Scalars & Vectors - OCR A-Level Physics
Worked Example
Resolving a pulling force
A rope pulls a box along a horizontal floor with a force of $F = 120 \text{ N}$ at $\theta = 25^\circ$ above the horizontal. Find the horizontal and vertical components of the pulling force.
A rope pulls a box along a horizontal floor with a force of $F = 120 \text{ N}$ at $\theta = 25^\circ$ above the horizontal. Find the horizontal and vertical components of the pulling force.
Diagram pending
Box on a horizontal floor. Rope drawn from a corner of the box up to the right, making angle $25^\circ$ with the floor. Force $F = 120 \text{ N}$ labelled along the rope. Dashed horizontal arrow labelled $F_x = F\cos\theta$ and dashed vertical arrow labelled $F_y = F\sin\theta$ shown as components of the rope force.
Will be replaced with a GeoGebra SVG in stream 2.
- The angle is measured from the horizontal floor, so the horizontal component is adjacent to $\theta$ and uses $\cos$; the vertical component is opposite $\theta$ and uses $\sin$.
- Substitute $F = 120 \text{ N}$ and $\theta = 25^\circ$.
$$F_x = F\cos\theta = 120 \times \cos 25^\circ = 120 \times 0.906 = 108.8 \text{ N}$$
$$F_y = F\sin\theta = 120 \times \sin 25^\circ = 120 \times 0.423 = 50.7 \text{ N}$$
- The horizontal component ($108.8 \text{ N}$) pulls the box forward along the floor and must overcome friction.
- The vertical component ($50.7 \text{ N}$) acts upward; it partially lifts the box and reduces the normal contact force, so $N = mg - F\sin\theta$.
- Pythagoras check: $\sqrt{F_x^2 + F_y^2} = \sqrt{108.8^2 + 50.7^2} = \sqrt{11837 + 2570} = \sqrt{14407} = 120 \text{ N}$. Matches the original force.
Common Mistake
MEDIUM
Wrong: Using $F_x = 120\sin 25^\circ$ because "horizontal feels like sin", or quoting the normal contact force as $N = mg$ (forgetting the rope lifts the box).
Right: The angle is measured from the horizontal, so $F_x = F\cos\theta$ (adjacent uses cos). The rope's vertical component reduces the normal force: $N = mg - F\sin\theta$.
Right: The angle is measured from the horizontal, so $F_x = F\cos\theta$ (adjacent uses cos). The rope's vertical component reduces the normal force: $N = mg - F\sin\theta$.
Examiner Tips and Tricks
- Always verify your components using Pythagoras: $\sqrt{F_x^2 + F_y^2}$ should equal the original force.
- If the question continues with friction or normal force, remember the rope's vertical component changes the effective weight on the floor.