MHB Solving 2.43: Drawing Diagrams & Vector Lengths

[Adhil]

I need help with 2.43
Is there any way to solve 2.43 without drawing the diagram again? And if I must draw another diagram, will I have to guess and check the length of the vectors since all the vectors are equal?

 

[Ackbach]

You don't need to re-draw the diagram, but you do need to do vector addition: component-wise. Take $\mathbf{T}_1$, for example. You're told it has a magnitude of $T$. Can you write down the components of $\mathbf{T}_1?$

 

[Adhil]

You don't need to re-draw the diagram, but you do need to do vector addition: component-wise. Take $\mathbf{T}_1$, for example. You're told it has a magnitude of $T$. Can you write down the components of $\mathbf{T}_1?$

The components of T1 will be T1x+T1y

 

[Ackbach]

The components of T1 will be T1x+T1y

Sorry, I meant can you write down the components of $\mathbf{T}_1$ in terms of its magnitude $T$ and its angle, call it $\theta_1=9^{\circ}$?

 

[Adhil]

Sorry, I meant can you write down the components of $\mathbf{T}_1$ in terms of its magnitude $T$ and its angle, call it $\theta_1=9^{\circ}$?

That would be T1=Tsin(9°)+Tcos(9°)

 

[Ackbach]

I would write it more like this (to emphasize its vector nature): $\mathbf{T}_1=\langle T \cos(9^{\circ}), T \sin(9^{\circ}) \rangle$, or $\mathbf{T}_1=T\cos(9^{\circ}) \,\hat{\mathbf{i}}+T\sin(9^{\circ}) \,\hat{\mathbf{j}}$.

Right, so then, you'd need to do this for the remaining three vectors. Once you've got all this written down, you can add the vectors together. What do you get?

 

[Adhil]

After removing T as a common factor I got 12500= T(3.27 + 2.05)

 

[Adhil]

After removing T as a common factor I got 12500= T(3.27 + 2.05)

Which means T=2349.6-lb...is this correct?
... Nvm, it's wrong

 

[Ackbach]

Which means T=2349.6-lb...is this correct?
... Nvm, it's wrong

Right. You need to set $T[\langle \cos(9^{\circ}), \sin(9^{\circ})\rangle + \dots + \langle \cos(51^{\circ}), \sin(51^{\circ})\rangle ] = 12500$. You have to find the magnitude of the vector part.

 

[Adhil]

Right. You need to set $T[\langle \cos(9^{\circ}), \sin(9^{\circ})\rangle + \dots + \langle \cos(51^{\circ}), \sin(51^{\circ})\rangle ] = 12500$. You have to find the magnitude of the vector part.

Okay thanks[emoji106]I get it now