Question

5. Etienne supports a rod as illustrated in the figure. What is the force exerted by each of his hands?

Answer 1

Given data:

* The length of the rod is L = 4 m.

* The mass of the rod is m = 10 kg.

* The distance of man's first hand from the end of of the rod is,

`\begin{gathered} x_1=30\text{ cm} \\ x_1=0.3\text{ m} \end{gathered}`

* The distance of man's second hand from the end of the rod is,

`\begin{gathered} x_2=100\text{ cm} \\ x_2=1\text{ m} \end{gathered}`

Solution:

The distance of man's first hand from the center of the rod is,

`\begin{gathered} d_1=(L)/(2)-x_1 \\ d_1=(4)/(2)-0.3 \\ d_1=2-0.3_{} \\ d_1=1.7\text{ m} \end{gathered}`

The distance of man's second hand from the center of the rod is,

`\begin{gathered} d_2=(L)/(2)-x_2 \\ d_2=(4)/(2)-1_{} \\ d_2=2-1 \\ d_2=1\text{ m} \end{gathered}`

As there is no rotation of rod takes place about the center of the rod, thus, the moment about the center due to the force from first hand is equal to the moment about the rod due to the force from second hand.

Thus,

`\begin{gathered} F_1* d_1=F_2* d_2 \\ F_1*1.7=F_2*1 \\ F_2=1.7F_1\ldots\text{.}(1) \end{gathered}`

The force exerted by the first hand is in the downward direction whereas the force exerted by the second hand is in an upward direction.

The diagrammatic representation of the given case is,

The net force exerted by the hands is handling the weight of the rod.

`F_2-F_1=mg`

where g is the acceleration due to gravity,

Substituting the known values,

`\begin{gathered} F_2-F_1=10*9.8 \\ F_2-F_1=98\ldots\text{.}(2) \end{gathered}`

From the equation (1) and (2),

`\begin{gathered} 1.7F_1-F_1=98 \\ 0.7F_1=98 \\ F_1=(98)/(0.7) \\ F_1=140\text{ N} \end{gathered}`

Similarly, the force exerted by the second hand is,

`\begin{gathered} F_2-F_1=98 \\ F_2=98+F_1 \\ F_2=98+140 \\ F_2=238\text{ N} \end{gathered}`

Thus, the force exerted by the first hand (right hand) is 140 N, and the force exerted by the second hand (left hand) is 238 N.