Question

Forces in equilibrium:

1) Given F1 = 1.47 N , where m1 = .150kg and $\theta$(1) = 210 (degrees)

Determine two more forces (magnitude and direction), which are at right angles to each other, but neither of them parallel to F(1), such that when they are applied, the system will be in equilibrium.

2) Given m1 = .140 kg at  $\theta$(1) = 80 (degrees), and m2 = .180 kg at  $\theta$(2) = 220 (degrees)

Determine the masses required in order that when two extra forces are applied, one parallel to line 1, and one parallel to line 2 in the diagram below, the whole system will be in equilibrium. Notice that line 1 and line 2 are perpendicular to each other, at specific angles shown.

2 lo 40° / 50° 22四 #娜瀞壽雛,乡興重

Answer 1

Solution:

Part A:

Given:

Fi 1.47N. mi 0.150kg, θι 210°

The two more forces say $F_{x}$ and $F_{y}$ which are at right angles to each other, but neither of them parallel to F(1), will be:

$F_{x}= F_{1} cos\theta$

181ηθ

Now,

F1.47 cos(210)

F, 1.47 *-0.866

1.27N

Negative sign shows the direction which is parallel to x axis.

F 1.47sin(210)

F,,-1.47 *-0.5

F,,-1.47 *-0.5

Fy-0.735N

Direction will be parallel to y axis.