IIT JEE , AIEEE Forum - Physics , Mechanics

srujana

Find T1 T2 T3 T4 & T5

swashata4iit

Is the length of hyportenuse is given

swashata4iit

Hi .................!~!!!!!!!!!!!!!!!!!!!!!!!!!!

I think this should be done like this

From FBD 1 (freeing from O)

T1 = 10 kg-f = 98 N

From FBD 2 (freeing from A)

By using lammi's theorem

T1/sin(180-@) = T2/sin(90+@) = T3/sin90

This implies that

T2 = T1cot@
T3 = T1/sin@

Now using FBD 3 (freeing from B)

T3/sin(180-@) = T4/sin180 = T5/sin@

It implies that
T5=T3= T1/sin@
T4 = 0

I cant find the @ Plz some body help

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netkid07

nicely explained

swashata4iit

Ok Srujana see this article for free body diagram

While studying mechanics, when we examine the forces acting on an object their are five "classic" types that are usually considered:

weight
normal
friction
tensions
applied forces

We use freebody diagrams to illustrate the magnitude and direction of all of the forces acting directly on a single object (usually represented by a rectangle). Consider a scenario in which a mass is being pulled across a table by a cord.

The weight vector begins at the object's center of mass and points towards the center of the earth.

A normal vector begins at the point of contact between the mass and its supporting surface. It is directed perpendicularly away from the surface and passes through the object's center of gravity.

Tensions are forces conducted along strings, ropes, and wires. They begin at the point of contact and point in the direction in which they are pulling.

Friction forces begin at the same point as the normal and act parallel to the sliding surface. They always oppose motion.

Applied forces is a catch-all, generic category encompassing any other interactions. In our current example, there are no generic applied forces.

If a force acts at an angle, then we usually work with its x- and y-components.

If an object is in static (at rest) or dynamic (constant velocity) equilibrium, then all of the forces acting on it are balanced.

The magnitude of the forces acting to the left equals the magnitude of the forces acting to the right.
The magnitude of the forces acting upwards equals the magnitude of the forces acting downwards.

In this case:

x: f = T cos ?

y:

+ T sin ? = mg

If the forces were not balanced, then the object would be accelerated in the direction of the unbalanced force. For example, using the same forces as in our previous example, if T cos ? were greater than f, then Newton's Second Law will allow us the ability to calculate the object's acceleration towards the right as it starts gaining speed.

net F = ma

T cos ? - f = ma (a > 0)

However, if T cos ? were less than f, then the object would still move towards the right but it would be losing speed.

net F = ma

T cos ? - f = ma (a < 0)

swashata4iit

Now for lammi's theorem refer to this

Statement: If three co-planers forces P, Q, R be in equilibrium and the angles between them be p (between Q n R), q (between P n R) and r (between P and Q) then

P/sinp = Q/sinq = R/sinr

For the proof see the diagram

I've also included the image explaining the angles between the strings.

Plz do vote my efforts

swashata4iit

And one more thing while drawing Free body diagram if a tension is involved (as we know it is bidirectional or nothin but a pull) always take the away vector or that sense of the tension which is going away from the point where from u r freeing the system (or drawing the FBD)

swashata4iit

Hope now it is clear 4 u

coolank2

The whole system is in equilibrium.

Hence, the equations will be .....

(1) T₁=mg

(2) At the lower point,

T₃.cos 37 = T₁

T₃.sin 37 = T₂

(3) At upper point,

T₅.cos 53 = T₃.cos 37

T₅.sin 53 = T₄ + T₃.sin 37

and we know that tan 37 = 4 / 3

Hence,

T₁ = 10g

T₂= 7.5g

T₃= 12.5g

T₄= 5.84g

T₅= 16.67g

Hope you get it .....

swashata4iit

The angle is not 37 and 53 As srujana told me the length of the dotted lines are 37 and 53. T5 and T3 belong to the same str. line. The angles are also same (see my diagram)

coolank2

Well in that case, the length of hypotnuse should be given.

T₄ = 0

And the angle be @, and hypotnuse be x.

Then T₃ = T₅ = T (say) ...... as same string.

Then we get,

T₁ = T.sin @ = 10g

T₂ = T.cos@

Hence, T = 9g/x

and T₂ = (

(8100 - x²))g / 9

This will do .........

But now also I think that acc to the figure, 37 and 53 are the angles ........

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