Find the tension in the two strings

In summary, the diagram shown is not a free body diagram. The long explanation is unnecessary. Forces are vectors, which means they always have both magnitudes and directions. Like all vectors, forces can be expressed in these components, which give the force's influence along the horizontal and vertical axes. The horizontal and vertical components of the same forces, T₁ and T₂, respectively, must then equate to zero. With the help of trigonometry, we can express T₁ₓ and T₂ₓ in terms of T₁ and T₂, respectively. If these are intended as final answers, you have forgotten the units and 3 decimal places is unjustifiable
  • #1
chwala
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Homework Statement
See attached;
Relevant Equations
Mechanics
This is an exam question - refreshing as usual.

1671443495984.png


Literature Overview;
In the free-body diagram shown above, we can see the horizontal and vertical components of the tension forces, T₁, and T₂. Forces are vectors, which means they always have both magnitudes and directions. Like all vectors, forces can be expressed in these components, which give the force's influence along the horizontal and vertical axes. T₁ₓ and T₂ₓ are the horizontal components of T₁ and T₂, respectively. On the other hand, T1y and T2y are the vertical components of the same forces, respectively. Since gravity acts on the object in the vertical axis, we need to consider the tension forces' vertical components for our summation of forces as follows:

##ΣF↑ = 0 = T_1 y + T_2 y + (-W)##

##T=W## where ##T=T_1 y + T_2 y##

##16 = T_1 \sin 40^0 + T_2 \sin 20^0##

We can also say that for the system to be in equilibrium, the object should not move horizontally or along the x-axis. Therefore, the horizontal components of T₁ and T₂ must then equate to zero. Also, with the help of trigonometry, we can express T₁ₓ and T₂ₓ in terms of T₁ and T₂, respectively:

##T_1 x = T_2 x##

##T_1 \cos 40^0 =T_2 \cos 20^0##

Therefore we shall have the simultaneous equation;

##16 = T_1 \sin 40^0 + T_2 \sin 20^0##

##T_1 \cos 40^0 =T_2 \cos 20^0##, Where;

##T_1=\dfrac{T_2 \cos 20^0}{\cos 40^0}##

Substituting on;

##16 = T_1 \sin 40^0 + T_2 \sin 20^0##, we shall have

##16=\dfrac {T_2 \sin 40^0 \cos 20^0}{\cos 40^0} + T_2 \sin 20^0##

##T_2(0.7885+0.342)=16##

##T_2 = 14.153## to 3 decimal places.

##T_1=\dfrac{14.153 ×\cos 20^0}{\cos 40^0}=17.361## to three decimal places.
Find mark scheme solution here;

1671445178033.png
Your insight is welcome.
 
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  • #2
chwala said:
In the free-body diagram shown above,
The diagram shown is not a free body diagram.

chwala said:
we can see the horizontal and vertical components of the tension forces, T₁, and T₂. Forces are vectors, which means they always have both magnitudes and directions. Like all vectors, forces can be expressed in these components, which give the force's influence along the horizontal and vertical axes. T₁ₓ and T₂ₓ are the horizontal components of T₁ and T₂, respectively. On the other hand, T1y and T2y are the vertical components of the same forces, respectively. Since gravity acts on the object in the vertical axis, we need to consider the tension forces' vertical components for our summation of forces as follows:
The long explanation is unnecessary. Look at the mark scheeme - there are no marks associated with the explanation. In an examination, this would cost you valuable time for zero benefit.

chwala said:
We can also say that for the system to be in equilibrium, the object should not move horizontally or along the x-axis.
This is irrelevant. The question tells you that the object is in equlibrium. (In any case, an object can be moving and still be in equlibrium - providing its velocity is constant.)

chwala said:
Therefore, the horizontal components of T₁ and T₂ must then equate to zero. Also, with the help of trigonometry, we can express T₁ₓ and T₂ₓ in terms of T₁ and T₂, respectively:
You don't need to say all that.

chwala said:
##T_1 x = T_2 x##
To be consistant with your approach to the vertical components, you should write
##T_1 x + T_2 x = 0##

chwala said:
##T_2 = 14.153## to 3 decimal places.
##T_1=\dfrac{14.153 ×\cos 20^0}{\cos 40^0}=17.361## to three decimal places.
If these are intended as final answers, you have forgotten the units and 3 decimal places is unjustifiable.

Your understanding is good but (IM O) you should make your answers far more concise.

And note that Lami's method is much simpler!
 
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  • #3
Steve4Physics said:
The diagram shown is not a free body diagram.The long explanation is unnecessary. Look at the mark scheeme - there are no marks associated with the explanation. In an examination, this would cost you valuable time for zero benefit.This is irrelevant. The question tells you that the object is in equlibrium. (In any case, an object can be moving and still be in equlibrium - providing its velocity is constant.)You don't need to say all that.To be consistant with your approach to the vertical components, you should write
##T_1 x + T_2 x = 0##If these are intended as final answers, you have forgotten the units and 3 decimal places is unjustifiable.

Your understanding is good but (IM O) you should make your answers far more concise.

And note that Lami's method is much simpler!
@Steve4Physics thanks...kindly note that the literature (copy pasted from google) is only for my reference and future review of my own posts...my interest is solely on the understanding of the concept as it has been long since attempting these questions...feedback noted though- thanks mate.
 
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  • #4
:chwa
chwala said:
@Steve4Physics thanks...kindly note that the literature (copy pasted from google) is only for my reference and future review of my own posts...my interest is solely on the understanding of the concept as it has been long since attempting these questions...feedback noted though- thanks mate.
My apologies @chwala. I thought you had written the solution. Do you understand it? Or is there something you are struggling with?
 
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  • #5
Steve4Physics said:
:chwa

My apologies @chwala. I thought you had written the solution. Do you understand it? Or is there something you are struggling with?
I understand it...

I am just but refreshing ...I majorly teach Maths ...though in my undergraduate degree i did Physics (minor)... always good to refresh on the concepts. An interesting area was Electromagnetism, Polymer physics...cheers mate.
 
  • #6
I would clarify in the solution that the units of weight and tensions are Newtons.
The hanging mass of 1.6 kg becomes 16 N in the initial equation without showing the mg step.
 
  • #7
Lnewqban said:
I would clarify in the solution that the units of weight and tensions are Newtons.
The hanging mass of 1.6 kg becomes 16 N in the initial equation without showing the mg step.
Yes, that's pretty straightforward; we use ##w=mg##. Where ##g=10## m/s^2.
 
  • #8
chwala said:
Yes, that's pretty straightforward; we use ##w=mg##.
Also, it may be useful to mention to the students that in practical terms, the angles could never be zero, because the magnitudes of the tensions would be infinite and either the rope/chain/cable or the anchors would fail.
 
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1. How do you calculate the tension in two strings?

The tension in two strings can be calculated using the formula T = (m1 + m2)g, where T is the tension, m1 and m2 are the masses of the objects connected by the strings, and g is the acceleration due to gravity.

2. What factors affect the tension in two strings?

The tension in two strings is affected by the masses of the objects connected by the strings, the length of the strings, and the angle at which the strings are being pulled.

3. Can the tension in two strings be greater than the weight of the objects?

Yes, the tension in two strings can be greater than the weight of the objects connected by the strings. This can happen when the strings are being pulled at an angle, resulting in a greater tension force.

4. How does the tension in two strings change if one of the strings is cut?

If one of the strings is cut, the tension in the remaining string will increase. This is because the weight of the object will now be supported by only one string instead of two.

5. How can the tension in two strings be reduced?

The tension in two strings can be reduced by decreasing the weight of the objects connected by the strings, increasing the length of the strings, or decreasing the angle at which the strings are being pulled.

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