Lesson 5, Topic 2
In Progress

# Condition of Equilibrium under the Action of Non-Parallel Coplanar Forces

Lesson Progress
0% Complete

The conditions of equilibrium under the action of non-parallel coplanar forces are:

1. Forces: The vector sum of all the forces acting on a body must be zero i.e.

(i) The algebraic sum of the horizontal components must be zero ∑Fx = 0

(ii) The algebraic sum of the vertical components of the forces must be zero ∑Fy = 0

2. Moments:  The algebraic sum of the moments of all the forces about any axis perpendicular to the plane of the forces must be zero.

### Example 1:

In the diagram below an object of mass 500kg is kept in equilibrium by three cables. Find the tension in each of the three cables. (g = 9.8ms-2)

Solution:

Tension in cable 3 can be found easily

T3 = mg

∴ T3 = 500 x 9.8 = 4900N (in the negative direction)

To find T1 and T2, since the object is in equilibrium ∑Fx = 0 and ∑Fy = 0

The horizontal component of T1 = T1X

The vertical component of T1 = T1Y

The horizontal component of T2 = T2X

The vertical component of T2 = T2Y

The horizontal component T1x = T1 cos 30º (Note: this is moving in the negative direction)

The vertical component T1y = T1 sin 30º

The horizontal component of T2x = T2 cos 30º

The vertical component of T2y = T2 sin 30º

Sum of the horizontal components ∑Fx;

= T2x – T1x = 0 (we subtract since T1x is moving in the negative direction)

T2 cos 30º – T1 cos 30º = 0

0.9397T2 – 0.8660T1 = 0 ……(i)

Sum of the horizontal components ∑Fy;

= T2y + T1y – T3 = 0

T2 sin 30º + T1 sin 30º – 4900 = 0

0.3420T2 + 0.5T1 – 4900 = 0 …….(ii)

We can then find T1 and T2 by substituting.

From equation (i)

0.9397T2 – 0.8660T1 = 0

0.9397T2 = 0.8660T1

T2 = $$\frac{0.866}{0.9397}\scriptsize T_1$$

T2 = 0.9216 T1 ….(iii)

substitute the value of T2 into equation (ii)

0.3420T2 + 0.5T1 – 4900 = 0 …….(ii)

0.3420(0.9216T1) + 0.5T1 – 4900 = 0

0.3152T1 + 0.5T1 = 4900

0.8152T1 = 4900

T1 = $$\frac{4900}{0.8152}$$

T1 = 6011N

To find T2 substitute the value of T1 into equation (iii)

T2 = 0.9216 T1 ….(iii)

T2 = 0.9216 x 6011

T2 = 5540N

### Example 2:

A body of mass of 10kg is kept in equilibrium by two ropes as shown in the diagram below. If one rope pulls the body in the horizontal direction and the other in a direction of 40º with the vertical, calculate the tension in each rope. (g = 10m/s2)

Solution:

The mass of 10k is equal to a force of 100N (F = mg = 10 x 10 = 100N)

To find T1 and T2, since the object is in equilibrium ∑Fx = 0 and ∑Fy = 0

The horizontal component of T1 = T1

The vertical component of T1 = 0 (no vertical component)

The horizontal component of T2 = T2X

The vertical component of T2 = T2Y

The horizontal component T1x = T1 (Note: this is moving in the negative direction)

The horizontal component of T2x = T2 cos 50º

The vertical component of T2y = T2 cos 40º

Sum of the horizontal components ∑Fx;

T2 cos 50º – T1 = 0

0.6427T2 = T1

or T1 = 0.6428T2

Sum of the vertical components ∑Fx;

T2 cos 40º – 100 = 0

T2 cos 40º = 100

∴ T2 = $$\frac{100}{cos40^{\circ}}$$

T2 = 130.54N

but

T1 = 0.6428T2

T1 = 0.6428 x 130.54

T1 = 83.909N

T1 = 83.91N

### Example 3:

A metal frame weighing 40N is hung on a nail using two ropes as shown. Find the tension in the rope.

Solution:

Due to symmetry, the tension in each rope will be the same. Because of this, the quickest way to solve this question is by simply writing an equilibrium equation in the y-direction: (there is no need resolving in x-dirextion)

Resolving in vertical direction, ∑Fy= 0

T cos60º + T cos60º – 40N = 0

2Tcos60º = 40N

Tcos60º = $$\frac{40}{2}$$

Tcos60º = 20

$$\scriptsize T \: \times \: \normalsize \frac{1}{2} \scriptsize = 20$$

$$\frac{T}{2} \scriptsize = 20$$

T = 40N

error: