Ropes pull at angles, handles push at angles. A force at an angle acts partly along each axis, so you split it with components and use the second law along each axis separately.
Key idea
Resolve every force into $x$- and $y$-components. Then $\Sigma F_x = ma_x$ and $\Sigma F_y = ma_y$ are two separate equations. A force at an angle changes both sums, including the normal force.
Why it works
The $x$ and $y$ directions are independent, as they were for adding vectors: the horizontal part of a pull can only change horizontal motion, and its vertical part only vertical motion. A rope pulling up at an angle lifts part of the object's weight, so the floor pushes up less, and friction ($\mu N$) gets smaller too.
- Draw the free-body diagram and split each slanted force into components.
- Write $\Sigma F_y = ma_y$ first (often $a_y = 0$) and solve for the normal force.
- Use that normal force for any friction: $f = \mu F_N$.
- Write $\Sigma F_x = ma_x$ and solve for the acceleration.
Worked examples
Worked example: Normal force with a slanted pull
A rope pulls a $20\ \text{kg}$ box along a level floor with $50\ \text{N}$ at $30^\circ$ above the horizontal. Find the normal force.
- Rope components: $50\cos 30^\circ \approx 43.3\ \text{N}$ forward, $50\sin 30^\circ = 25\ \text{N}$ up.
- $\Sigma F_y = 0$: $F_N + 25 - 196 = 0$, so $F_N = 171\ \text{N}$.
Answer: $171\ \text{N}$
Worked example: Adding friction
Same box, with $\mu_k = 0.2$. Find its acceleration.
- $f_k = \mu_k F_N = 0.2 \times 171 = 34.2\ \text{N}$ (not $0.2 \times 196$).
- $\Sigma F_x = 43.3 - 34.2 = 9.1\ \text{N}$, so $a = \dfrac{9.1}{20} \approx 0.46\ \text{m/s}^2$.
Answer: about $0.46\ \text{m/s}^2$
Worked example: Pushing down at an angle
You push a $15\ \text{kg}$ mower with $60\ \text{N}$ along its handle, $35^\circ$ below the horizontal. Find the normal force.
- The push's vertical part is $60\sin 35^\circ \approx 34.4\ \text{N}$ down.
- $$F_N = mg + 34.4 = 147 + 34.4 \approx 181\ \text{N}.$$
Answer: about $181\ \text{N}$
Common mistake
Forgetting that a rope pulling up at an angle also lowers the normal force (and that a push down at an angle raises it), then using $\mu mg$ for friction. Check: write the $y$-equation before the $x$-equation, and find $F_N$ from it first.