This is so annoying me

Gughanath

Another problem I cant solve

A box of mass 6kg lies on a rough plane inclined at an angle of 30 to the horizontal. The box is held in equilibrium by means of a horizontal force of magnitude P newtons.
The coefficent of friction between the box and the plane is 0.4.

Find the normal reaction exerted on the body by the plane.

The box is in limiting equilibrium and on the point of moving up the plane.

Nylex

You know that in limiting equilibrium, the frictional force is [itex]F = \mu R[/itex], where R is the normal reaction.

The box is in equilibrium, but if it were to move, it would move up the plane. That should tell you in which direction the forces on the box are (or rather, if they're in the same direction, or in opposite directions).

Draw a diagram, those help loads in A Level Maths!

Edit: you don't even need all that info to find the normal reaction. Resolve the weight into components parallel and perpendicular to the slope and you can get the normal reaction from that.

Gughanath

but the force P (which is unknown at the moment) will also need to be taken into consideration. If I resolved into perpendicular components it becomes Pcos60 +6gcos30 = R?

Nylex

Ahh, sorry. I didn't see that horizontal force bit. Yes, you're right about that.

Gughanath

I included P in my equations and then elinated it. But I dont get the right answer. What would be your suggested method?

Nylex

Post your working, then we can see where you've gone wrong (if anywhere!). You're not using Edexcel books are you? Sometimes they get the answers wrong.

Gughanath

Resolving perpendicualr to the plane:

R = Pcos60 + 6gcos30

Resolving parallel to the plane:

Fr (friction) = Pcos30 - 6gcos60

is this right so far?

Doc Al

Write the equations representing the equilibrium condition for forces parallel and perpendicular to the plane. You'll get two equations and two unknowns (P and the Normal reaction); solve for N.

Nylex

Yep, looks ok to me.

Gughanath

and Fr (the friction will also be in the equation which is the problem.

Nylex

Why is the frictional force a problem?

Doc Al

But the friction can be written in terms of R. Since it is just about to move, the static friction will equal its maximum value [itex]F_r = \mu R[/itex], as was pointed out by Nylex in his first post in this thread.

Gughanath

ok. I just did everything again. And I found out that I went wrong with the sins and cos during the calcualtions. My value for R comes out to be 88.3N which is correct.

Gughanath

Yes I used Fr = uR

Andrew Mason

The last point tells you that the actual friction force is the maximum static friction force (ie. exactly the coefficient of static friction multiplied by the normal force). The components of force along and perpendicular to the plane are:

Parallel to plane:

(1) [tex]mgsin(\theta) + Pcos(\theta) + \mu_s N = 0[/tex]


Perpendicular:

(2) [tex]mgcos(\theta) + Psin(\theta) = N[/tex]

Substitute N from (2) into (1) to solve for P in terms of mg and [itex]\theta[/itex]:

[tex]mgsin(\theta) + Pcos(\theta) + \mu_s (mgcos(\theta) + Psin(\theta)) = 0[/tex]

[tex]mgsin(\theta) + \mu_smgcos(\theta) + P(cos(\theta) + \mu_ssin(\theta)) = 0[/tex]

[tex]P = \frac{-(mgsin(\theta) + \mu_smgcos(\theta))}{cos(\theta) + \mu_ssin(\theta)}[/tex]

Substitute that into (2) to get N.

AM

Gughanath

yep...thanx....works