Solved Problem on Dynamics

The fiure represents a system is composed of an elevator of mass M and a mass of mass m. The elevator is hanging by a rope that passes through a fixed pulley and comes to the operator's hands, the rope and the pulley are supposed frictionless and have a negligible mass. The operator pulls the rope and rises with constant acceleration a along with the elevator. It is assumed known M, m, a, and g. Determine the force that the elevator exerts on the operator.

Problem data:

Mass of man: m;
Mass of elevator: M;
Acceleration of the system: a;
Acceleration due to gravity: g.

Solution

Assuming the direction of the acceleration is positive, we draw a free-body diagram. We find the forces on each of them and we apply Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m \vec{a}} \tag{I} \end{gather} \]

Man (Figure 1):

\( {\vec F}_{gH} \): gravitational force on the man;
\( \vec{T} \): tension force applied by the man on the rope;
\( \vec{N} \): force of reaction of the elevator on man (force to determine).

In this problem, there is the only movement in the vertical direction, applying the expression (I) to the man

\[ \begin{gather} T+N-F_{gH}=ma \tag{II} \end{gather} \]

Figure 1

Elevator (Figure 2):

\( {\vec F}_{gE} \): gravitational force on the elevator;
\( \vec{T} \): tension force due to the pull that the man gives on the rope;
\( \vec{N} \): force of action of the man on the elevator.

Applying the expression (I) to the elevator

\[ \begin{gather} T-N-F_{gE}=Ma \tag{III} \end{gather} \]

Figure 2

Equations (II) and (III) can be written as a system of two equations to two variables, T and N

\[ \begin{gather} \left\{ \begin{array}{l} T+N-F_{gH}=ma \\ T-N-F_{gE}=Ma \end{array} \right. \end{gather} \]

as we want to get the value of the force of reaction N that the elevator does in man, we will subtract the second equation from the first equaton

\[ \begin{gather} \frac{ \begin{aligned} \cancel{T}+N-F_{gH}=ma \\ \text{(-)}\qquad \cancel{T}-N-F_{gE}=Ma \end{aligned} } {0+2N-F_{gH}+F_{gE}=ma-Ma}\\ N=\frac{F_{gH}-F_{gE}+ma-Ma}{2} \tag{IV} \end{gather} \]

The gravitational force is given by

\[ \bbox[#99CCFF,10px] {F_{g}=mg} \]

the gravitational force of man is given by

\[ \begin{gather} F_{gH}=mg \tag{V} \end{gather} \]

the gravitational force of the elevator is given by

\[ \begin{gather} F_{gE}=Mg \tag{VI} \end{gather} \]

substituting the expressions (V) and (VI) into expression (IV)

\[ \begin{gather} N=\frac{mg-Mg+ma-Ma}{2}\\ N=\frac{g(m-M)+a(m-M)}{2} \end{gather} \]

factoring (m−M)

\[ \bbox[#FFCCCC,10px] {N=\frac{(m-M)(g+a)}{2}} \]

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