Equilibrium

When all the forces acting on a body is balanced or the net force is zero, the object is said to be in its equilibrium state. Whether the object is moving or not, it could be in equilibrium as long as the sum of all the forces acting on it is equal to zero.

\(\sum F=0\)

In case the forces are on different directions, the sum of all the components of the forces must also be zero.

\(\sum F_x=0\)

\(\sum F_y=0\)

The sign of the forces varies on its direction. Positive force means the force is pointing towards positive direction and negative sign to negative direction.

To solve equilibrium problems, the following steps are followed:

  1. Draw the Free-Body diagram of all the forces acting on the object.
  2. In case the direction of the forces are in an angle, resolve the x- and y-components of the forces.
  3. Get the sum of the forces along each axis.
  4. Solve for the unknown force.

There are two types of equilibrium – static equilibrium and dynamic equilibrium. When an object is at rest, it is said to be in its state of static equilibrium. When a body is moving, but with constant velocity, it is in the state of dynamic equilibrium.


Example 1.

Manuel is sitting on the chair. If his mass is \(35-kg\), how much force does the table exerted on him?

Solution 1.

FBD:

 


Solution 2.

\(\sum F_x=0\)


Solution 3.

\(\sum F_y=0\)

\(F_N-F_g=0 \implies N=F_g\)

\(F_g\) is equal to the weight of Manuel which can be solved as

\(F_g=W=mg=(35\;kg)(9.8m/s^2)=343\;N\)

Substituting the value of \(F_g\) to the equation, we get \(F_N\) as

\(F_N=343\;N\).


Example 2.

Solve for the mass of a loaded cart that is pushed horizontally on the floor by a force of \(200\;N\). The box is moving with constant velocity and the coefficient of kinetic friction is \(0.35\).


Solution 1.

FBD:


Solution 2.

\(\sum F_x=0\)

\(F_a-F_f=0 \implies F_a=F_f\)

Since the applied force is \(200.0 \;N\), it follows that the friction force is also \(200.0 \;N\).


Solution 3.

Solve for the normal force using the formula for friction \(F_f=\mu N\).

\(F_f=\mu N \implies N=\frac{F_f}{\mu}\)

 

\(N=\frac{200.0\;N}{0.35}=571.43\;N\)


Solution 4.

\(\sum F_y=0\)

\(F_N-F_g=0 \implies F_N=F_g\)

Since \(F_N\) equals \(571.43\;N\), \(F_g\) is also equals \(571.43\;N\).


Solution 5.

Use the value of \(F_g\) to solve for the mass of the cart.

\(F_g=W=mg\)

 

\(W=mg \implies m=\frac {W}{g}\)

 

\(m=\frac{571.43\;N}{9.8\;m/s^2}=58.31\;kg\)

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