- #1

Leo Liu

Gold Member

- 226

- 89

- Homework Statement:
- This is a statement.

- Relevant Equations:
- This is an equation.

For question b, the official solution sets up a non-inertial coordinate on the block and writes out the following two equations:

$$\begin{cases}

\begin{align*}

f\cos(\theta)+N\sin(\theta)-mg=0 \qquad \hat\jmath

\\

N\cos(\theta)+f\sin(\theta)=ma \qquad\quad\;\;\, \hat\imath

\end{align*}

\end{cases}$$

The questions above are true because the horizontal acceleration of the block must be the same as the acceleration of the wedge to keep the former on the latter.

However, if I try to solve for the normal force using (2), I will obtain ##N=\frac{mg}{\mu \sin(\theta) + \cos(\theta)}## which shows that the magnitude of the normal force is independent to the acceleration.

This counterintuitive conclusion confuses me-- if you push it with a greater horizontal force, the block will be pinned harder to the ground because the block needs a stronger force to accelerate in lockstep with the wedge, and therefore creates a larger normal force. Could someone explain me why this is the case? Thanks in advance :).

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