Question

To find the magnitude of counterbalancing force contributed by three given forces.

Short answer

The magnitude of counterbalancing force is \(26.1\;{\rm{N}}\).

Step-by-step solution

Concept of counterbalancing Force and formula used

When two items of equal weight, power, or influence act in opposition to each other, a counterbalance is a weight or force that balances or offsets the other. The items are said to be in counterbalance at this point.

Given:

Magnitude of\(3\)forces as\(25\;{\rm{N}},12\;{\rm{N}}\), and\(4\;{\rm{N}}\)with angles as\({100^^\circ },{100^^\circ }\), and\({90^^\circ }\)respectively.

Formula used:

Consider the expression for component form of vector in\(x\)-axis.

\({\bf{v}} = |{\bf{v}}|{\bf{i}}.......\left( 1 \right)\)

Consider the expression for component form of vector in\(xy\)plane.

\({\bf{v}} = |{\bf{v}}|\cos \theta {\bf{i}} + |{\bf{v}}|\sin \theta {\bf{j}}........(2)\)

Consider the expression for component form of vector in\(z\)-axis.

\({\bf{v}} = |{\bf{v}}|{\bf{k}}.......(3)\)

Write the expression for magnitude of vector\({\bf{a}} = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}(|{\bf{a}}|)\).

\(|{\bf{a}}| = \sqrt {a_1^2 + a_2^2 + a_3^2} .......(5)\)

Calculate magnitude of counterbalancing force

Consider the object is located at origin and three forces are located on\(x,y\)and\(z\)- axes.

Set the first force with magnitude\(25\;{\rm{N}}\)to lie on\(x\)-axis.

Find the force vector of first force\(\left( {{{\bf{v}}_{\bf{1}}}} \right)\)by using equation (1).

\({{\bf{v}}_1} = 25{\bf{i}}{\rm{N}}\)

Set the second force with magnitude\(12\;{\rm{N}}\)to lie on\(xy\)plane

Find the force vector of second force\(\left( {{{\bf{v}}_2}} \right)\)by using equation (1).

\(\begin{aligned}{l}{{\bf{v}}_2} = 12\cos {100^^\circ }{\bf{i}} + 12\sin {100^^\circ }{\bf{j}}{\rm{N}}\\ = - 2.0838{\bf{i}} + 11.8177{\bf{j}}{\rm{N}}\end{aligned}\)

Set the third force with magnitude\(4\;{\rm{N}}\)to lie on\(z\)-axis.

Find the force vector of third force\(({{\bf{v}}_3})\)by using equation (1).

\({{\bf{v}}_3} = 4{\bf{k}}\;{\bf{N}}\)

Substitute\(25{\bf{i}}{\rm{N}}\)for\({{\bf{v}}_1}, - 2.0838{\bf{i}} + 11.8177{\bf{j}}{\rm{N}}\)for\({{\bf{v}}_2}\), and\(4{\rm{kN}}\)for\({{\bf{v}}_3}\)in equation (4),

\(\begin{aligned}{l}{\bf{v}} = 25{\bf{i}}{\rm{N}} + ( - 2.0838{\bf{i}} + 11.8177{\bf{j}}){\rm{N}} + 4{\bf{k}}{\rm{N}}\\ = 25{\bf{i}} - 2.0838{\bf{i}} + 11.8177{\bf{j}} + 4{\bf{k}}{\rm{N}}\\ = (25 - 2.0838){\bf{i}} + 11.8177{\bf{j}} + 4{\bf{k}}{\rm{N}}\\ = 22.9162{\bf{i}} + 11.8177{\bf{j}} + 4{\bf{k}}{\rm{N}}\end{aligned}\)

Here,

\({\bf{v}}\)is resultant counterbalance vector by three forces.

Find the magnitude of\({\bf{v}}(|{\bf{v}}|)\)by using equation (5).

\(\begin{aligned}{l}|{\bf{v}}| = \sqrt {{{(22.9162\;{\rm{N}})}^2} + {{(11.8177\;{\rm{N}})}^2} + {{(4\;{\rm{N}})}^2}} \\ = \sqrt {525.152 + 139.658 + 16} \;{\rm{N}}\\ = \sqrt {680.81} \;{\rm{N}}\\ = 26.09\;{\rm{N}}\\|{\bf{v}}| \simeq 26.1\;{\rm{N}}\end{aligned}\)

Thus, the magnitude of counterbalancing force is \(26.1\;{\rm{N}}\).