You are Here: Home

# Tension and pulleys mechanics Tweet

Maths and statistics discussion, revision, exam and homework help.

Announcements Posted on
TSR launches Learn Together! - Our new subscription to help improve your learning 16-05-2013
IMPORTANT: You must wait until midnight (morning exams)/4.30AM (afternoon exams) to discuss Edexcel exams and until 1pm/6pm the following day for STEP and IB exams. Please read before posting, including for rules for practical and oral exams. 28-04-2013
1. Tension and pulleys mechanics
knowledge of M1 only:

Two particles P and Q of mass 2 kg and 3 kg respectively are connected by a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a rough inclined plane. The plane is inclined to the horizontal at an angle of 30°. Particle P is held at rest on the inclined plane and Q hangs freely on the edge of the plane with the string vertical and taut. Particle P is released and it accelerates up the plane at 2.5m s−2. Find

a the tension in the string,

b the coefficient of friction between P and the plane,

c the force exerted by the string on the pulley.

c)

I've drawn a diagram shown below, trying to show the tension and the force coming out of the pulley:

Now for working I tried to imagine that I have the force F already and i'm trying to find T, so:

Fcos30 = T
Fsin30 = T
Fcos30 + Fsin30 = 2T
F(cos30+sin30) = 2T
F = 2T/(cos30+sin30) and subbing T in there gives me the answer of 32.068... which is incorrect.

I took a look at the solutions and they are much different to mine:

F=2Tcos30
= 43.8cos30
= 37.9

and they ahve also drawn a different diagram, any help?
2. Re: Tension and pulleys mechanics
It may have been simpler to draw a force triangle, of the forces at the pulley.

Also, when you resolve the force F, you should do so orthogonally.

You should have had F.cos30 - vertically down.
And F.sin30 - going horizontally - not in the same direction as the T going down the plane.

The solution is :-

Horizontal Forces--------Vertical Forces
T.cos30 = F.cos60-------T + T.sin30 = F.cos30

Either of the above two equations can be solved for F.
3. Re: Tension and pulleys mechanics
(Original post by steve10)
It may have been simpler to draw a force triangle, of the forces at the pulley.

Also, when you resolve the force F, you should do so orthogonally.

You should have had F.cos30 - vertically down.
And F.sin30 - going horizontally - not in the same direction as the T going down the plane.

The solution is :-

Horizontal Forces--------Vertical Forces
T.cos30 = F.cos60-------T + T.sin30 = F.cos30

Either of the above two equations can be solved for F.
What do you mean by orthogonally? Also, how do you draw a force triangle, I haven't been taught this. I see how you done it though, helps thanks (i'm assuming by orthogonally you mean at right angles).
Last edited by Whizzkid1; 06-07-2012 at 18:57.
4. Re: Tension and pulleys mechanics
Orthogonallly means at right angles, yes.

You haven't been taught to draw a force triangle, no? No problem, you'll probably come accross it later. To draw a force triangle, let's take the forces acting at the pulley and draw a force diagram, then use that to create our force triangle.

The force triangle has the same forces as the force diagram, and in the same directions.
Take the force F from the force diagram and draw a line parallel to that force on another part of your paper. Mark it with F. Similarly draw a line parallel to the force T2 (a vertical line) and draw that on your paper so that it cuts (intersects with) the line of the force F that you just drew. Mark it with T2.
Then draw a line parallel to the force T1 so that it cuts the other two lines that you have just drawn and so will give you a triangle.
Bear in mind that the arrows indicating the directions of the forces should always follow each other around the triangle. (This helps you to construct the triangle and comes in very handy if your system involves 5 or 6 forces, say, acting at a point and you have to draw a polygon of forces, not just a triangle!)
This is a vector triangle of forces. The lines of the triangle are vectors and represent the forces F, T1 and T2 in both size and direction. The lengths of the sides of your vector force triangle represent the sizes of the forces in your force diagram.
The angles between the forces in your force diagram can be used to give the angles in your force triangle. e.g. in your original force diagram , from your question, you would have had T1 = T2 = T and the angle alpha = 60 degrees.
This would have made your force triangle an isosceles triangle with two small angles = 30 degrees and a large angle = 120 degrees.
If you were now to drop a perpendicular from the vertex of the large angle to the mid-point of the force F, then you would end up with a 30-60-90 triangle with hypotenuse = T, adjacent side = F/2 and included angle = 30 giving cos(30) = (F/2)/T, or F = 2Tcos(30) = T.sqrt(3).

You would use a force diagram to solve a statics problem by resolving horizontally and vertically, then equating forces in the horizontal and vertical directions.
You would use a force triangle to solve a statics problem by using trigonometry to solve for the unknowns.

When resolving a force, normally you always do this orthogonally. (See bottom of attachment)