M1 Jan 2003 Help!!!

hey guys

need help on question 5:

5) A box of mass 1.5 kg is placed on a plane which is inclined at an angle of 30 degrees to the horizontal. The coefficient of friction between the box and the plane is 1/3. The box is kept in equilibrium by a light string which lies in a vertical plane containing a line of greatest slope of the plane. The string makes an angle of 20 degrees with the plane. The box is in limiting equilibrium and is about to move up the plane. The tension in the string is T newtons. The box is modelled as a particle.

Find the value of T.

10 MARKS

I have the formulas when resolving perpendicular and parallel to the plane:

Tsin20 + R = 1.5gcos30 (perpendicular)

Tcos20 = F + 1.5gsin30 (parallel)

F = 1/3 R

But then when I solve for T, I dont get the right answer which should be 11.0 Newtons.

How do I get the 11 N?

Thanks very much
hatty341

PS - There's a diagram of the question beneath

untitled.JPG11.2KB

Reply 1

insparato

15

Tsin20 + R = 1.5gcos30 (perpendicular)

Tcos20 = F + 1.5gsin30 (parallel)

F = 1/3 R

Tcos20 = 1/3R + 1.5gsin30

Tcos20 -1.5gsin30 = 1/3R

Tsin20 + R = 1.5gcos30

R = 1.5gcos30 - Tsin20

Tcos20 -1.5gsin30 = 1/3(1.5gcos30 - Tsin20)

3tcos20 - 4.5gsin30 = 1.5gcos20 - Tsin20

3tcos20 + Tsin20 = 1.5gcos 20 + 4.5gsin30

t(3cos20 + sin20) = 1.5gcos 20 + 4.5gsin30

t = 1.5gcos 20 + 4.5gsin30 / (3cos20 + sin20)

t = 13.81348153 + 22.05 / (3.161098006)

t = 11.35

t = 11N to the nearest newton.

I've used the equations given the algebra ive used seems to be alright hmm.

Reply 2

generalebriety

16

insparato

Tcos20 -1.5gsin30 = 1/3(1.5gcos30 - Tsin20)

3tcos20 - 4.5gsin30 = 1.5gcos20 - Tsin20

Bang.

Reply 3

generalebriety

16

Continuing,

3Tcos20 - 4.5gsin30 = 1.5gcos30 - Tsin20
3Tcos20 + Tsin20 = 1.5gcos30 + 4.5gsin30
T(3cos20 + sin20) = 1.5gcos30 + 4.5gsin30
T = (1.5gcos30 + 4.5gsin30) / (3cos20 + sin20)
T = 34.78... / 3.16...
T = 11.00... = 11.0 N. :smile:

Reply 4

insparato

15

Bah i hate computers.

Too many sin's and cos's

Reply 5

AJC98

1

I don't know what the rules are on dragging up 8 year old threads so sorry if I'm stepping on anyone's toes, but how did you get those original equations? The

Tsin20 + R = 1.5gcos30 (perpendicular)

Tcos20 = F + 1.5gsin30 (parallel)

Thanks :smile:

Reply 6

dinodazer

1

Original post by AJC98

I don't know what the rules are on dragging up 8 year old threads so sorry if I'm stepping on anyone's toes, but how did you get those original equations? The

Tsin20 + R = 1.5gcos30 (perpendicular)

Tcos20 = F + 1.5gsin30 (parallel)

Thanks :smile:

Don't worry, I just did this in my Mechanics course too :biggrin:

To get the perpendicular forces, you use the only two forces that have a vertical component, ie The Tension in the string and the normal reaction force. Therefore you can say that:

RN = 1.5gcos30 - Vertical component of Tension

So:

RN = 1.5gcos30 - Tsin20

Then rearrange to get:

Tsin20 + RN = 1.5gcos30

Then for parallel it's saying that the total force is the horizontal component of the Tension, minus the horizontal component of the weight so:

F=Tcos20 - 1.5gsin30

Rearrange to get:

Tcos20 = F + 1.5gsin30

Hope this helps! :biggrin:

Reply 7

AJC98

1

Original post by dinodazer

Don't worry, I just did this in my Mechanics course too :biggrin:

To get the perpendicular forces, you use the only two forces that have a vertical component, ie The Tension in the string and the normal reaction force. Therefore you can say that:

RN = 1.5gcos30 - Vertical component of Tension

So:

RN = 1.5gcos30 - Tsin20

Then rearrange to get:

Tsin20 + RN = 1.5gcos30

Then for parallel it's saying that the total force is the horizontal component of the Tension, minus the horizontal component of the weight so:

F=Tcos20 - 1.5gsin30

Rearrange to get:

Tcos20 = F + 1.5gsin30

Hope this helps! :biggrin:

Aaaahhh ok thanks!!

M1 Jan 2003 Help!!!

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