- #1

chwala

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- Homework Statement:
- See attached;

- Relevant Equations:
- Mechanics

This is an exam question - refreshing as usual.

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

##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.

Your insight is welcome.

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;**

Your insight is welcome.

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