Determine is box of unknown mass is sliding down a ramp?

[Jojo96]

Homework Statement

A box of textbooks at unknown mass rests on a loading ramp that makes an angle α = 16o with the horizontal. The coefficient of static friction is 0.30, and kinetic friction is 0.25. Then the box is released. (a) Will it start sliding down the ramp? (a) Find the normal force on the box (b) If it slides down then find the acceleration of the box

2. Relevant
∑F=ma
ƒ=µn

The Attempt at a Solution

Forces in x-direction:
w-ƒstatic=0
w=ƒstatic
mgsin(16)=µn
y: n-w=0
n=w
n=mgcos(16)
------
mgsin(16)=µmgcos(16)
sin(16)=µcos(16)
sin(16)/cos(16)=µ
But already know µ so this doesn't help me...
[/B]

 

[Potatochip911]

By assuming ##\sum F_x=0## you're stating that the block isn't moving in the x direction. Try using ##\sum F_x=ma_x##

 

[drvrm]

3. The Attempt at a Solution
Forces in x-direction:
w-ƒstatic=0
w=ƒstatic
mgsin(16)=µn
y: n-w=0
n=w
n=mgcos(16)
------
mgsin(16)=µmgcos(16)
sin(16)=µcos(16)
sin(16)/cos(16)=µ
But already know µ so this doesn't help me...

with the ramp at your angle theta - if you calculate the force driving the box then perhaps it will come out less than the frictional force opposing the tendency to slide down- as it will be mu. M.g .cos (angle)

 

[Jojo96]

The only force that is driving it down is the weight. Nothing is pulling the box. Also, what would I use for acceleration in the x direction for F=ma?

 

[cnh1995]

The only force that is driving it down is the weight. Nothing is pulling the box. Also, what would I use for acceleration in the x direction for F=ma?

Component of weight along the slope is pulling the box down. Check if it is sufficient to overcome static friction.

 

[drvrm]

The only force that is driving it down is the weight. Nothing is pulling the box. Also, what would I use for acceleration in the x direction for F=ma?

along the ramp the force acting downward will be component of Mg - i.e. Mg sin(16)- but your body can not move with this pull along the ramp as frictional force is larger - if you increase the angle of ramp then only it can move at an angle Theta such that Mgsin(theta) exceeds
(Coeff. of friction XMg Cos(theta)) which is opposing the motion acting along the ramp in upward direction.