find T1 and T2, if a 60 N load is hanged and in equilibrium position.​

Find T1 And T2, If A 60 N Load Is Hanged And In Equilibrium Position.

Answers

Answer 1

Answers:

T1 = 59.5677879151077 N

T2 = 45.7583538175564  N

The values are approximate. Round the values however you need to.

=========================================================

Explanation:

Refer to the diagram below. I have added red and blue lines to break up each tension vector into the x and y components. For tension T1 in red, we have x1 and y1 as the horizontal and vertical components respectively. Note that x1 points to the left and y1 points upward.

For T2 in blue, we have x2 point to the right and y2 point upward. The horizontal components x1 and x2 must be equal, and point in opposite directions, so that we have horizontal balance. Otherwise, the object would accelerate to one side rather than be stationary.

Applying a bit of trig, we can say

cos(angle) = adjacent/hypotenuse

cos(45) = (x1)/(T1)

x1 = T1*cos(45)

and also

sin(angle) = opposite/hypotenuse

sin(67) = (x2)/(T2)

x2 = T2*sin(67)

----------------------

Again, the x1 and x2 vectors must be equal in magnitude, allowing us to say this

x1 = x2

T1*cos(45) = T2*sin(67)

T1 = T2*sin(67)/cos(45)

which we'll use later.

----------------------

Doing a bit more trig set up, we know that

sin(45) = (y1)/(T1)

y1 = T1*sin(45)

and

cos(67) = (y2)/(T2)

y2 = T2*cos(67)

Adding those two y components together will yield the total upward force that the ropes apply

y1+y2 = T1*sin(45)+T2*cos(67)

y1+y2 = (T2*sin(67)/cos(45))*sin(45)+T2*cos(67)

y1+y2 = T2*( sin(67)*sin(45)/cos(45) + cos(67) )

y1+y2 = T2*( sin(67)*tan(45) + cos(67) )

y1+y2 = T2*( sin(67)*1 + cos(67) )

y1+y2 = T2*( sin(67) + cos(67) )

This combined upward pull must be equal to 60 newtons or else we wouldn't have vertical equilibrium. If the upward pull was smaller than the downward pull, then the object would be falling. If the other way around, then the object would be pulled upward at some accelerated speed (think of a spring action).

So we'll set y1+y2 equal to 60 and solve for T2

y1+y2 = 60

T2*( sin(67) + cos(67) ) = 60

T2 = 60/( sin(67) + cos(67) )

T2 = 45.7583538175564

That value is approximate. The units for T2 are in newtons. This is the amount of pull or force along the T2 line in the upper right direction.

Side note: make sure your calculator is in degree mode.

----------------------

The last thing to do is to compute T1

T1 = T2*sin(67)/cos(45)

T1 = 45.7583538175564*sin(67)/cos(45)

T1 = 59.5677879151077

That value is approximate as well. The units are also newtons here. This represents the amount of pull in the upper left direction as the diagram shows.

----------------------

Finding such values for T1 and T2 is useful to determine if a rope can handle a certain stress load. Let's say we had a rope that was rated for only 40 newtons. Anything beyond 40 N for this hypothetical rope would mean the rope breaks. We found T1 and T2 beyond this threshold which would indicate that such a rope isn't up to the task of holding this object up (at least not with the angles shown in the diagram).

Find T1 And T2, If A 60 N Load Is Hanged And In Equilibrium Position.

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