The 6.4 N weight is in equilibrium under the influence of the three forces actin

Question

The 6.4 N weight is in equilibrium under the influence of the three forces acting on it. The F1 force acts from above on the left at an angle of 34degree with the horizontal. The F2 force acts from above on the right at an angle of 56degree with the horizontal. The 6.4 N force acts straight down. Note: F1 and F2 are not to scale. What is the magnitude of the force F2, as shown in the figure above? Answer in units of N What is the magnitude of the force F1, as shown in the figure above? Answer in units of N

Explanation / Answer

I am taking horizontal as x axis and vertical as y axis. Take components of F1 & F2 in x-y direction and write the equilibrium equation:

   Fx = 0

F1* cos (34) - F2 * cos(56) =0

   F1 = 0.6745 F2      ------(1)

Fy = 0                       ( Taking upward direction as positive)

F1* sin (34) + F2 * sin(56) - 6.4 = 0

0.6745 F2 *sin (34) + F2 * sin(56) =6.4       ( use equation (1))

F2 = 5.3058 N (Answer)

So F1 = 0.6745 *5.3058

F1 = 3.5787 N (Answer)

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