Find the magnitude and direction of the net force acting on each of the objects in the following diagrams:

Question

Find the magnitude and direction of the net force acting on each of the objects in the following diagrams:

^ 60 N

/

/

/

_________/___ ) 40° (the curve is 40 degree)

| |___________> 40 N

| |

|___________ |

Given:

Ask:

Formula:

Solution:

Answer:

PLEASE GIVE THE 5. sorry for the drawing, hope you understand it.

Answer 1

Break the 60 N vector into two components like this:

So the force in the x-direction is about 46.963 N and the force in the y-direction is about 38.567 N.

Now, combine the two force in the x-direction: 46.963 N + 40 N = 85.963 N. Now, use the Pythagorean Theorem and put them back together like this:

85.963² + 38.567² = x² so x = √(85.963² + 38.567²) the answer will be about 94 N but do the math to get the exact answer)

Calculate the angle using θ = tan^-1(38.567/85.963) which will be in the range of 24° (do the math to get the exact answer)

Given: Two forces acting on the object, one is 40 N on the horizontal, one is 60 N at 40° above the horizontal.

Ask: What is the resultant force?

Solution (the math and pictures shown above)

Answer: "x" N at x° above the horizontal (fill in the "x's" by doing the math as explained)

Answer 2

60N at 40 degrees

40N at 0 degrees

angles formed with the horizontal

construct a parallelogram and calculate the magnitude and direction of the longer diagonal

sides of the parallelogram are 40,60,40 and 60

opposite angle also = 40 degrees

other 2 angles of the parallelogram are 180-40 = 140 degrees each

use law of cosines to find the diagonal = d

d^2 = a^2 +b^2 -2abCosD, D is the angle opposite side d

d^2 = 60^2 +40^2 -2(60)(40)Cos140

d= sqrd^2 = about 94.218 Newtons

from a rough sketch you know it had to be substantially larger than 60 N

and less than 60+40=100 N

when you get the longer diagonal's length, look at the resulting lower triangle and find its angles

use the law of sines

sin140/d = sinA/40 where A is the direction of the resultant force, d= longer diagonal length

sinA =40sin140/d= about 0.2729

A = sin^-1(40sin140/d)=arcsin.2729 =about 15.84 degrees

= about 16 degrees = angle of the longer diagonal with the horizontal

magnitude of the resultant force is the length of the longer diagonal

its direction=A= is the angle of one of the triangles formed by the diagonal cutting the parallelogram in half

this is basically a trigonometry problem using laws of sines and cosines

that comes up in most precalculus courses and introductory physics and engineering courses

no guarantees the above calculations are error free, but they look about right for the expected range of values