Force needed to stop a block sliding down a 30 degree ramp

Expert intro: On a smooth ramp, there is no friction to help you. The only thing trying to move the block is the component of its weight down the slope, so the balancing force must match that exactly.

Detailed walkthrough

The block is on a smooth ramp, so the only force parallel to the ramp is the component of its weight.

1) Find the weight

Mass = $20\,\text{kg}$, and take $g=9.8\,\text{N/kg}$: $W=mg=20\times 9.8=196\,\text{N}$

2) Resolve weight along the slope

For an incline of $30^\circ$, the component down the ramp is $W\sin 30^\circ=196\times 0.5=98\,\text{N}$

3) Force needed to prevent motion

To keep the block from moving, apply an equal force up the ramp: $\boxed{98\,\text{N}}$

That force must act parallel to the ramp, directed upward.

💡 Pitfall guide

Do not use $\cos 30^\circ$ for the parallel component. With the angle given from the horizontal, the downhill component is always $mg\sin\theta$. Mixing that up gives the wrong answer by a noticeable margin.

🔄 Real-world variant

If the ramp were rough, friction could share part of the job, so the required applied force would be smaller. If the angle changed to $\theta$, the needed force would be $mg\sin\theta$ on a smooth incline.

🔍 Related terms

inclined plane, component of weight, equilibrium