(III) You are on a pirate ship and being forced to walk the plank (Fig. 9–68). You are standing at the point marked C. The plank is nailed onto the deck at point A, and rests on the support 0.75 m away from A. The center of mass of the uniform plank is located at point B. Your mass is 65 kg and the mass of the plank is 45 kg. What is the minimum downward force the nails must exert on the plank to hold it in place?

Short Answer

Expert verified

The minimum downward force that the nails must exert on the plank to hold it in place is \(2352\;{\rm{N}}\).

Step by step solution

01

Meaning of torque

The torque value, mathematically, can be determined by multiplying the value of the applied force and the distance of the force from a pivoted point.

02

Given information

Given data:

The mass of the person is \(M = 65\;{\rm{kg}}\).

The mass of the plank is \(m = 45\;{\rm{kg}}\).

03

Free body diagram of the plank

The free body diagram of the plank can be drawn as:

Here, \({F_{{\rm{nails}}}}\) is the force exerted on the nails and \({F_{{\rm{pivot}}}}\) is the force exerted on the pivot point.

04

Evaluation of the minimum downward force that the nails must exert on the plank to hold it in place

Take torques about the pivot point, with the clockwise torques, as positive.

\(\begin{array}{c}\sum \tau = 0\\\left\{ \begin{array}{l}mg\left( {0.75\;{\rm{m}}} \right)\cos \theta + Mg\left( {2.25\;{\rm{m}}} \right)\cos \theta \\ - {F_{{\rm{nails}}}}\left( {0.75\;{\rm{m}}} \right)\cos \theta \end{array} \right\} = 0\\{F_{{\rm{nails}}}} = \frac{{mg\left( {0.75\;{\rm{m}}} \right)\cos \theta + Mg\left( {2.25\;{\rm{m}}} \right)\cos \theta }}{{\left( {0.75\;{\rm{m}}} \right)\cos \theta }}\\{F_{{\rm{nails}}}} = \frac{{mg\left( {0.75\;{\rm{m}}} \right) + Mg\left( {2.25\;{\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}{F_{{\rm{nails}}}} = \frac{{\left( {45\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {0.75\;{\rm{m}}} \right) + \left( {65\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2.25\;{\rm{m}}} \right)}}{{\left( {0.75\;{\rm{m}}} \right)}}\\{F_{{\rm{nails}}}} = 2352\;{\rm{N}}\end{array}\)

Thus, the minimum downward force that the nails must exert on the plank to hold it in place is \(2352\;{\rm{N}}\).

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(I) Calculate the mass m needed in order to suspend the leg shown in Fig. 9–47. Assume the leg (with cast) has a mass of 15.0 kg, and its CG is 35.0 cm from the hip joint; the cord holding the sling is 78.0 cm from the hip joint.

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