Question

\[ \begin{array}{l} -T_{1} \cos (37)+T_{2} \cos (53)=0 \\ T_{1} \sin (37)+T_{2} \sin (53)-T_{3}=0 \\ T_{3}=m g, m=2 k g \\ T_{3}=2 * 9.81=19.6 N \\ \vec{T}_{3}=0 \vec{\imath}-19.6 \vec{\jmath} N \end{array} \] Find $ T_{1} $ and $ T_{2} $

          \[
\begin{array}{l}
-T_{1} \cos (37)+T_{2} \cos (53)=0 \\
T_{1} \sin (37)+T_{2} \sin (53)-T_{3}=0 \\
T_{3}=m g, m=2 k g \\
T_{3}=2 * 9.81=19.6 N \\
\vec{T}_{3}=0 \vec{\imath}-19.6 \vec{\jmath} N
\end{array}
\]
Find \( T_{1} \) and \( T_{2} \)

-T1cos (37)+T2cos (53)=0

T1sin (37)+T2sin (53)-T3=0

T3=m g, m=2 k g

T3=2 * 9.81=19.6 N
T⃗3=0 -19.6 N

Find T1 and T2

Added by Sara A.

Applied Calculus for the Managerial, Life, and Social Sciences

Soo T. Tan 10th Edition

Chapter 8

Instant Answer

Solved by Expert Sikandar Baig

Step 1

6 = 0$$ Simplifying and solving for $T_2$, we get: $$T_2 = \frac{19.6}{\sin(53) + \cos(37)\sin(53)} \approx 23.5 N$$ Substituting this back into the first equation, we can solve for $T_1$: $$T_1 = \frac{T_2 \cos(53)}{\cos(37)} \approx 16.8 N$$ Show more…

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