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Find the tension in the ropes shown in the figure if the supported object weighs 600N.

Find The Tension In The Ropes Shown In The Figure If The Supported Object Weighs 600N class=

Sagot :

LammettHash

By Newton's second and third laws, we have the following net forces at

  • point A:

[tex]\sum F_x = -T_1 \cos(60^\circ) + T_2 \cos(60^\circ) = 0[/tex]

[tex]\sum F_y = T_1 \sin(60^\circ) + T_2 \sin(60^\circ) - 600 \,\mathrm N = 0[/tex]

Together, these equations tell us [tex]\boxed{T_1=T_2\approx 346 \,\mathrm N}[/tex].

  • point B:

[tex]\sum F_x = T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5 = 0[/tex]

[tex]\sum F_y = T_3 \sin(20^\circ) - T_2 \sin(60^\circ) = 0[/tex]

  • point C:

[tex]\sum F_x = -T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]

[tex]\sum F_y = T_4 \cos(20^\circ) - T_1 \sin(60^\circ) = 0[/tex]

Combining the horizontal force equations for points B and C and using the fact that [tex]T_1=T_2[/tex] gives

[tex](T_3 \cos(20^\circ) - T_2 \cos(60^\circ) - T_5) + (-T_4  \cos(20^\circ) + T_1 \cos(60^\circ) + T_5) = 0 + 0[/tex]

[tex]\implies (T_3 - T_4) \cos(20^\circ) = 0 \implies T_3=T_4[/tex]

Then combining the vertical force equations for B and C, we find

[tex](T_3 \sin(20^\circ) - T_2 \sin(60^\circ)) + (T_4 \sin(20^\circ) - T_1 \sin(60^\circ)) = 0 + 0[/tex]

[tex]\implies 2 T_3 \sin(20^\circ) - 2 T_1 \sin(60^\circ) \implies T_3 = \dfrac{\sin(60^\circ)}{\sin(20^\circ)} T_1[/tex]

so that [tex]\boxed{T_3 = T_4 \approx 877 \,\mathrm N}[/tex]

Lastly solve for [tex]T_5[/tex] using either horizontal force equation for B or C.

[tex]-T_4 \cos(20^\circ) + T_1 \cos(60^\circ) + T_5 = 0[/tex]

[tex]\implies T_5 = \left(\dfrac{\sin(60^\circ)}{\sin(20^\circ)} - \cos(60^\circ)\right) T_1 \implies \boxed{T_5 \approx 651\,\mathrm N}[/tex]

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