- #1
marsupial
- 45
- 2
Homework Statement
A block of mass ## M_1 ## is on an inclined plane (the plane is inclined at 30 degrees), and is attached with a light cord over a light, frictionless pulley to another mass ## M_2 ##. ## M_1 ## is fixed at 5.00kg, and ## M_2 ## can be varied.
When ## M_2 = 3.20 kg ##, ## M_1 ## just begins to slide up the slope, acclerating at ## 0.20 m/s^2 ##.
Determine the coefficients of static and kinetic friction between ## M_1 ## and the plane.
Homework Equations
## F_{fr_static} = \mu (static) F_N ##
## F_{fr_kinetic} = \mu (kinetic) F_N ##
F = ma
The Attempt at a Solution
After resolving the components of mg into x, and y: ## F_N = y = 5.00gCos30 = 42.435244 N ##
The component of mg in the x direction: mgsin30 = 24.5N
To determine the coefficient of static friction, there must be no net horizontal force. I am not sure of this, but I am wondering if I just ignore the pulley and the mass at the end to calculate the static friction coefficient? But then that assumes that it will sit on the plane without sliding down, and I am not sure if I should make that assumption.
If I did make that assumption, and ignored the pulley for the moment, then:
## F_{fr} - 5.00gsin30 = 0 --> \mu F_N = 24.5 --> \mu (static) = 24.5/42.43 = 0.577 ##
To determine the coefficient of kinetic friction:
## net F_x = 3.20 * 0.20 ##
## F_{tension} - F_{fr_kinetic} - F_{mg-xcomponent} = 3.20 * 0.2 ##
## F_{tension} = 3.20g ##
## F_{fr_kinetic} = (F_{tension} - F_{mg-xcomponent})/ (3.20 * 0.2) ##
## F_{fr_kinetic} = (3.20g - 24.5)/ (3.20 * 0.2) = 10.71875 ##
## \mu (kinetic) = F_{fr_kinetic} / F_N = 10.71875/ 42.435244 = 0.253 ##
I'm not sure in particular about the coefficient of static friction.