- #1
Net Force of A: Diagram Analysis
AI Thread Summary
To find the net force on object A, the relevant equations include the force between charges FAB=KqAqB/r², and the components of the forces Fx=Fcosθ and Fy=Fsinθ. The total force can be calculated using the formula FR=sqrt(Fx² + Fy²) and determining the angle θ with θ=tan⁻¹(Fy/Fx). Participants are encouraged to calculate the vector forces F_AB and F_AC, then sum these vectors to obtain the total force acting on A. Providing the calculations will help clarify the process.
- #2
BvU
Science Advisor
Homework Helper
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Well, what is your question?
## \ ##
- #3
MatinSAR
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I think you want to find Fnet A ...
Tell me which part you need explanation.
Tell me which part you need explanation.
- #4
berkeman
Admin
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You need to find the vector forces ## F_{AB} ## and ## F_{AC} ## and sum those force vectors to get the total force on A. Can you show us your work on that?Spotify said:
Here we know that if block B is going to move up or just be at the verge of moving up
##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ##
Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point
Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A.
I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Figure 1 Overall Structure Diagram
Figure 2: Top view of the piston when it is cylindrical
A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N.
Figure 3: Modifying the structure to incorporate a fixed internal piston
When I modify the piston...
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