A force of 800 N is exerted on a bolt A as shown. Determine the horizontal and vertical components of the force.
We will use the trigonometric functions of the angle α = 35°.
Fx=−F×cosα=−(800 N)cos35°=−655 NF_x = -F \times cos α = -(800 \;N) cos 35° = -655 \;NFx=−F×cosα=−(800N)cos35°=−655N
Fy=+F×sinα=+(800 N)sin35°=+459 NF_y = +F \times sin α = +(800\;N) sin 35° = +459 \;NFy=+F×sinα=+(800N)sin35°=+459N
The vector components of F are:
Fx=−(655 N)iF_x = -(655 \;N)iFx=−(655N)i
Fy=+(459 N)jF_y = +(459 \;N)jFy=+(459N)j
We can write F as
F = -(655 N)i + (459 N)j
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