Q2DQ - University Physics with Modern Physics

Question

“In general, the normal force is not equal to the weight.” Give an example in which these two forces are equal in magnitude, and at least two examples in which they are not.

Step-by-Step Solution

Answer

When the elevator moves upward or downward with constant velocity, then the normal force is equal to the weight of the man.

When the elevator is speeding upwards, the normal force is not equal to the weight.

When the elevator is speeding downwards, the normal force is not equal to the weight.

Step 1: Significance of the force.

The Force is described as an agent that mainly changes the motion of an object. The force has both the direction and also the magnitude.

Step 2: Determination of an example of the forces that are equal in magnitude.

The free-body diagram of an elevator is drawn below;

 

 

According to the above diagram, if an elevator moves with a constant velocity, then the acceleration of the whole system will be zero that is $a_y=0$. The summation of the forces in the y direction is zero.

 

The equation of the summation of the forces in the y direction is expressed as:

 

$$\begin{gathered} \underset{ }{ \sum }{F}_y=0 \\ {F}_n-mg=m{a}_y \end{gathered}$$ 

Here, $\sum _{ }{F}_y$ is the summation of the forces in the y direction, $F_n$ is described as the normal force, m is the mass of the person, g is the acceleration due to gravity and $a_y$ is the acceleration of the system.

Substitute 0 for $a_y$ in the above equation.

$$\begin{gathered} F_n-mg=m{a}_y \\ F_n=mg \end{gathered}$$ 

Hence, the normal force is equal to the force exerted by the man.

Thus, when the elevator moves upward or downward with constant velocity, then the normal force is equal to the weight of the man.

Step 3: Determination of the example when the normal force is not equal to weight.

The diagram of the elevator speeding up has been drawn below:

 

 

According to the above figure, if the elevator is speeding up, then the acceleration of the system will be constant. The summation of the forces in the y direction is zero.

The equation of the summation of the forces in the y direction is expressed as:

$$\begin{gathered} \sum _{ }{F}_n=0 \\ {F}_n-mg=m{a}_y \\ F_n=mg+m{a}_y \end{gathered}$$

 

Hence, the above equation states that $F_n>mg$.

Thus, when the elevator is speeding upwards, the normal force is not equal to the weight.

Step 4: Determination of another example when the normal force is not equal to weight.

The diagram of the elevator speeding downwards has been drawn below:

 

 

According to the above figure, if the elevator is speeding downwards, then the acceleration of the system will be negative. The summation of the forces in the y direction is zero.

The equation of the summation of the forces in the y direction is expressed as:

 

$$\begin{gathered} \sum _{ }{F}_y=0 \\ {F}_n-mg=m{a}_y \\ F_n=mg-m{a}_y \end{gathered}$$

Hence, the above equation states that $F_n<mg$.

Thus, when the elevator is speeding downwards, the normal force is not equal to the weight.