A) \[{{F}_{3}}-{{F}_{1}}-{{F}_{2}}\]
B) \[\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\]
C) \[\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}-F_{2}^{2}}\]
D) \[{{F}_{3}}-{{F}_{1}}+{{F}_{2}}\]
Correct Answer: B
Solution :
The FBD of the loop is as shown Therefore, force on QP will be equal and opposite to sum of forces on other sides. Thus, \[{{F}_{QP}}=\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\] Alternative: \[{{F}_{4}}\sin \theta ={{F}_{2}}\] \[{{F}_{4}}\cos \theta =({{F}_{3}}-{{F}_{1}})\] \[\therefore \]\[{{F}_{4}}=\sqrt{{{({{F}_{3}}-{{F}_{1}})}^{2}}+F_{2}^{2}}\]You need to login to perform this action.
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