- #1
- 1,752
- 143
Can someone tell me if I did this right?
An object of mass m is hung from a rope that passes over a pulley at the top of a ramp and is attached to a block of mass M.
(a) Assume m is large enough that the block accelerates up the ramp. Find an expression for the block’s acceleration.
(b) From your result for part (a), determine the minimum ratio m/M such that the block, once moving, accelerates up the ramp.
(the diagram is simple, M is on the incline plane and m is hanging straight down)
[tex]a = \frac{Force}{mass}[/tex]
[tex]a=\frac{Mg sin\theta + \mu Mg cos \theta -mg}{M}[/tex] (answer for part A)
Set the force equal to 0
[tex]Mg sin \theta + \mu Mg cos \theta - mg = 0[/tex]
[tex]Mg sin \theta + \mu Mg cos \theta = mg [/tex]
factor out Mg
[tex]Mg(sin \theta + \mu cos \theta) = mg[/tex]
the g's cancel
[tex]m = M(sin \theta + \mu cos \theta)[/tex]
[tex]m/M = sin \theta + \mu cos \theta[/tex]
An object of mass m is hung from a rope that passes over a pulley at the top of a ramp and is attached to a block of mass M.
(a) Assume m is large enough that the block accelerates up the ramp. Find an expression for the block’s acceleration.
(b) From your result for part (a), determine the minimum ratio m/M such that the block, once moving, accelerates up the ramp.
(the diagram is simple, M is on the incline plane and m is hanging straight down)
[tex]a = \frac{Force}{mass}[/tex]
[tex]a=\frac{Mg sin\theta + \mu Mg cos \theta -mg}{M}[/tex] (answer for part A)
Set the force equal to 0
[tex]Mg sin \theta + \mu Mg cos \theta - mg = 0[/tex]
[tex]Mg sin \theta + \mu Mg cos \theta = mg [/tex]
factor out Mg
[tex]Mg(sin \theta + \mu cos \theta) = mg[/tex]
the g's cancel
[tex]m = M(sin \theta + \mu cos \theta)[/tex]
[tex]m/M = sin \theta + \mu cos \theta[/tex]