A 5 500-N weight is held suspended in equilibrium by two cables.Cable 1 applies a horizontal force to the right of the object and has a tension,T₁.Cable 2 applies a force upward and to the left at an angle of 40.0° to the negative x-axis and has a tension,T₂.What is the tension,T₁? A)4.6E+3 N B)6.6E+3 N C)4.2E+3 N D)8.6E+3 N E)1.1E+4 N

Question

Quizplus · Accepted Answer

The tension in Cable 1 (T1) can be found by analyzing the forces in the horizontal direction. Since the system is in equilibrium, the horizontal component of the tension in Cable 2 (T2) must equal the tension in Cable 1. The horizontal component of T2 can be calculated using trigonometry, specifically by multiplying the tension in Cable 2 by the cosine of the angle (cos(40°)). Given that the weight is 5500 N and it is in equilibrium, the vertical component of T2 must also balance the weight. By solving these equations simultaneously, one can find the value of T1, which is 6.6E+3 N.

A 5 500-N Weight Is Held Suspended in Equilibrium by Two