Find the magnitude of the horizontal and vertical components of a force of 465 newtons at a bearing of 19° from the horizontal. Round to the nearest whole number.

Question

horizontal component: 440 N, vertical component: 151 N
	horizontal component: 440 N, vertical component: 25 N
	horizontal component: 151 N, vertical component: 314 N
	horizontal component: 151 N, vertical component: 440 N

Answer 1

The horizontal component of force is equal to the force multiplied by the cosine of the angle from the horizontal:

F_h = Fcos(θ)

The vertical component of force is equal to the force multiplied by the sine of the angle from the horizontal:

F_v = Fsin(θ)

Thus:

F_h = (465 N) x cos(19^o)

F_h = 439.67 N

F_h = 440 N (rounded)

F_v = (465 N) x sin(19^o)

F_v = 151.39 N

F_v = 151 N (rounded)

Answer 2

Draw a picture!!

The force is a vector pointing to the right, with the angle to the horizontal being 19 degrees

There is a right triangle with the stated force as the hypotenuse. The components of the force:

Vertical = 465 * sin 19

Horizontal = 465 * cos 19