Is Fgy the Same as FN in a System with Two Masses and Friction?

Click For Summary
The discussion revolves around the confusion regarding the normal force (FN) and gravitational force component (Fgy) in a system with two masses and friction. Participants clarify that Fgy is not the same as FN, emphasizing the need for a proper force diagram to understand the forces acting on the first mass (m1). The calculation for Fgy is presented, but the expression is deemed incorrect, prompting the suggestion to draw a force diagram for clarity. A missing diagram that previously illustrated the problem is mentioned, indicating it may have been deleted by the original poster. Understanding the relationship between the forces requires visual representation and correct application of trigonometric functions.
danielsmith123123
Messages
26
Reaction score
4
Please do not delete parts of your post after you have received help on schoolwork problems
Homework Statement
what is the acceleration and tension of this system
Coefficient of kinetic friction = 0.5
“” “ static “ = 0.6
Mass1 = 20kg
Mass2= 12kg
Relevant Equations
F= ma
Fg = mg
I am just confused on how to find the normal force/ FN of the first object. My classmates are saying Fgy is the exact same as Fn but I don’t get why

Fgy= Fg sin theta
Fgy= (20)(9.81) (sin35)
Fgy= 112.5

Fgy = FN
 
Physics news on Phys.org
Your classmates are wrong. Having said that, your expression is incorrect. Draw a force diagram for m1 and you will see why. Hint: sine = opposite/hypotenuse; cosine = adjacent/hypotenuse.
 
kuruman said:
Your classmates are wrong. Having said that, your expression is incorrect. Draw a force diagram for m1 and you will see why. Hint: sine = opposite/hypotenuse; cosine = adjacent/hypotenuse.
Ok, thank you
 
Is there a diagram that goes with this problem?
 
berkeman said:
Is there a diagram that goes with this problem?
There was, but now it's gone. It showed two masses, m1 on an incline attached to a rope over a pulley to a m2 hanging straight down.
 
Thanks @kuruman -- I looked in the post history and didn't see a figure. OP must have deleted it after you helped them.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
View full post »

Similar threads

Forces help (friction, tension, etc.)
Replies
2
Views
1K
Frictional force acting on mass
Replies
5
Views
3K
Pulley with 2 Blocks on an Inclined Plane. Help please
Replies
3
Views
12K
Acceleration on a pulley system with 3 blocks
Replies
3
Views
5K
Measuring the coefficient of friction
Replies
7
Views
3K
Friction and tensions with two blocks on a string on an incline
Replies
4
Views
17K
Two connected masses on separate inclines
Replies
20
Views
3K
How Do You Calculate Acceleration on an Inclined Plane with Friction?
Replies
2
Views
2K
Static & Kinetic Friction + An Applied Force on an Inclined Ramp
Replies
13
Views
4K
How does the angle of the handle affect the frictional force on the mower?
Replies
4
Views
3K