M1 - friction and equilibrium

Would somebody like to help me with these questions??

1) A parcel of mass 1kg is placed on a rough plane which is inclined at 30 degrees to the horizontal. The coefficient of friction between the parcel and the plane is 0.25. Fnid the force that must be applied to the parcel in a direction parallel to the plane so that:
a) the parcel is just prevented from sliding down the plane
b) the parcel is just on the point of moving up the plane

2) A box of mass 6kg is placed on a rough plane which is inclined at 45 degrees to the horizontal. The coefficient of friction between the box and the plane is 0.5. Find the horizontal force that must be applied to the box so that:
a) the box is just prevented from sliding down the plane
b) the box is just on the point of moving up the plane

Reply 1

mizfissy815

I'm guessing this is from Heinmann book...what's Ex. and question number?

Reply 2

turtle2

mizfissy815

I'm guessing this is from Heinmann book...what's Ex. and question number?

Nope, it's not from a book...it was a sheet that I got given.

Reply 3

massimo

on point of slipping down is when the downward force is at its greatest so you take away the horizontal compenent of force paralell to the plane.

r=1gcos30+xsin30
f=1gsin30-xcos30 then multiply by 0.25 and you should get x

when its moving up the plane is when the F is at its smallest so you take away the paralell force from the weight and add the horizontal paralell compenent as it would be going up so heres your second equation

r=1gcos30+xsin30 x0.25
f=-1gsin30+xcos30

and the other one is the same principle have a go at that one and if you cant do it come back on here or pm me and dont forget the pos rep either :wink:

thanks

Reply 4

StudentTazzy

Point of moving down the plane
The key point here is that the force acting down the plane has to equal the Fmax force (the maximum force of friction that can be applied).
When the force down the plane is greater than this then the object moves down the plane. When Fmax is greater than the force acting down the plane than friction holds the object still.

The Fmax force = the coeff of friction X R.

R = mgcos30 = 8.49 (3s.f.)
Coeff of Friction = 0.25
Fmax therefore = 2.1225

The force down the plane is mgsin30 = 4.9
Without any external force the object would move down the plane. You have to calculate the force acting up the plane so that 4.9 - this force = Fmax.

Rearranging this gives you: 4.9 - Fmax = the force.

Hopes this helps.
Give the other question ago using similar principles, remember drawing a clear diagram with all forces labelled will realy help. If you're having real problems PM me and I'll help you futher.

Good Luck, Tazzy :smile: