Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
A block on an accelerating wedge
Reply to thread
Message
[QUOTE="Leo Liu, post: 6371834, member: 660839"] [B]Homework Statement:[/B] This is a statement. [B]Relevant Equations:[/B] This is an equation. [ATTACH type="full" alt="119d7e6a6edde9801eb08ef273ef81f9.png"]266851[/ATTACH] For question b, the official solution sets up a non-inertial coordinate on the block and writes out the following two equations: $$\begin{cases} \begin{align*} f\cos(\theta)+N\sin(\theta)-mg=0 \qquad \hat\jmath \\ N\cos(\theta)+f\sin(\theta)=ma \qquad\quad\;\;\, \hat\imath \end{align*} \end{cases}$$ [ATTACH type="full" alt="1595632949641.png"]266798[/ATTACH] 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 :). [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
A block on an accelerating wedge
Back
Top