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Mechanics, Friction help

A sack of weight 600N is being loaded onto a truck up a ramp inclined at 18 degrees to the horizontal. A rope attached to the sack is held at an angle of 25 degrees to the ramp. The coeffeicient of friction between the sack and the ramp in 0.3.
What tension in the rope is needed

a) To prevent the sack from sliding down the ramp
b) to pull the sack up at a steady speed

I was able to get
Tcos(23) = F + 600sin(18)

and

600cos(18) = Tsin(23) + R

The equation im trying to use is flim=uR
but if i cant work out R, then i cant work out flim either

So i was hoping someone could help me get past just this 1 step

Reply 1

ghostwalker

Original post by Notsocleverstudent

Looks like you've resolved parallel to the ramp. Which way is friction acting, given it's on the point of slipping down?

and

600cos(18) = Tsin(23) + R

The equation im trying to use is flim=uR
but if i cant work out R, then i cant work out flim either

So i was hoping someone could help me get past just this 1 step

As you're asked for the tension needed, it's in limiting equilibrium and flim is F

So just use F=uR

And you then have 3 equations and 3 unknowns.

PS is the angle 23 or 25?

Reply 2

Notsocleverstudent

Original post by ghostwalker

Looks like you've resolved parallel to the ramp. Which way is friction acting, given it's on the point of slipping down?

Friction is acting against the driving force which if you draw on a diagram, the friction would be acting down the slope(ramp)
And the problem is, I get this far but without knowing either T or F, I'm kind of stuck

Original post by ghostwalker

As you're asked for the tension needed, it's in limiting equilibrium and flim is F

So just use F=uR

And you then have 3 equations and 3 unknowns.

PS is the angle 23 or 25?

Sorry, it's 25, my head kind of dazed off :frown:

And, the problem i get with using Flim = uR is that I can't work out R because the Tension T is acting both vertically so without T I can't work out R and vice versa

And was there a 3rd equation I'm not aware of :s

Thank you so far

Reply 3

ghostwalker

Original post by Notsocleverstudent

No, friction acts against the direction in which the body is tending to move. If the mass is moving up, friction acts down, and if the mass is about to slip down, friction acts upwards, so to speak.

That is the third equation F=uR.

Reply 4

Notsocleverstudent

Original post by ghostwalker

No, friction acts against the direction in which the body is tending to move. If the mass is moving up, friction acts down, and if the mass is about to slip down, friction acts upwards, so to speak.

Thank you. I actually didn't see it from that perespective. I'll draw a new diagram, and see if I can get further with that :smile:

That is the third equation F=uR.

So just to be sure, the 3 equations are:

Tcos(23) = F + 600sin(18)
600cos(18) = Tsin(23) + R
Flim = uR

And, if I rearrange the 1st 2, into the 3rd one, I would get

Tcos(23)-600sin(18) = 0.3 x 600cos(18)-Tsin(23)

Right so far? :/

Reply 5

ghostwalker

Original post by Notsocleverstudent

As I said in my first post, which direction is friction acting? Your first equation is incorrect.

Regarding your 3rd equation, don't forget Flim = F here.