Question 2.165: The rope CE exerts a 500-N force T on the hinged door. (a) E......
The rope CE exerts a 500-N force T on the hinged door.
(a) Express T in terms of components.
(b) Determine the vector component of T parallel to the line from point A to point B.
Step 1:
To find the tension vector T, we first need to calculate the position vector r_CE, which is given as (0.2i + 0.2j - 0.1k) m.
Step 2:
Next, we can calculate T by multiplying the magnitude of T (500 N) with the unit vector in the direction of r_CE, denoted as r_CE/|r_CE|. This unit vector is found by dividing r_CE by its magnitude.
Step 3:
After performing the calculations, we find T to be equal to (333i + 333j - 167k) N.
Step 4:
To find the component of T parallel to AB, we need to calculate the unit vector in the direction of AB, denoted as e_AB. This is found by dividing the position vector r_AB (-0.15i + 0.2k) by its magnitude.
Step 5:
Then, we can calculate the parallel component of T, denoted as T_p, by taking the dot product of e_AB and T, and multiplying it with e_AB.
Step 6:
By performing the calculations, we find T_p to be equal to (200i - 267k) N.
In summary, the tension vector T is equal to (333i + 333j - 167k) N, and the parallel component of T, T_p, is equal to (200i - 267k) N.
Final Answer
We have
\begin{aligned}&r _{C E}=(0.2 i +0.2 j -0.1 k )~m\\&T =(500~N ) \frac{ r _{C E}}{\left| r _{C E}\right|}=(333 i +333 j -167 k )~N\end{aligned}(a) T = (333i + 333j – 167k) N
(b) We define the unit vector in the direction of AB and then use this vector to find the component parallel to AB.
\begin{aligned}& r _{A B}=(-0.15 i +0.2 k )~m \\& e _{A B}=\frac{ r _{A B}}{\left| r _{A B}\right|}=(-0.6 i +0.8 k ) \\& T _p=\left( e _{A B} \cdot T \right) e _{A B}=([-0.6][333 N ]+[0.8][-167 N ])(-0.6 i +0.8 k ) \\& T _p=(200 i -267 k )~N\end{aligned}
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