Question

Two forces act on the hook. Determine the magnitude of the resultant force.

Answer 1

Answer:

Net Force in X-direction = 200cos30 + 500cos70 = 200*.866+500*.342 = 344.2N

Net Force in Y-direction = 200sin30 + 500sin70 = 200*.5+500*94 = 570 N

The net-force is sqrt(344.2^2 + 570^2) = 665.86 in the direction of tan-1(Fy/Fx)

= tan-1(570/344.2)

= tan-1(1.656)

= 58.87

Answer 2

Answer:

Solution

Draw the components as follows. (Note that the diagrams are not to scale)

Two forces act on the hook

Draw the vector components tail to tail as follows.

Two forces act on the hook

Use cosine law to figure out the resultant force F_{R}F

R

F_{R}=\sqrt{500^2+200^2-(2)(500)(200)(\cos140^0)}F

R = 5002+2002 −(2)(500)(200)(cos1400 )

F_{R}=665.7NF

R =665.7N