Referring to Fig.1.5, compute the coefficient of friction at which the tendency of the body to slide and the tendency to topple due to the applied force are equal.Please show me step by step and explain it. Thank you! 87F,1.5 mAPivot point0.1 mFIGURE 1.5A force applied to an erect person.直 078°F AARARpri scnFBhomeendFBdn 6d5. 0.1 mPivot pointFIGURE 1.5 A force applied to an erect person.usclesThe opposite restoring torque Tw due to the person's weight isT, =W×0.1m(1.2)WAssuming that the mass m of the person is 70kg, his weight W isingW =mg=70 × 9.8 = 686 newton (N)(1.3)g Tip-Toe(Here g is the gravitational acceleration, which has the magnitude 9.8m/sec².)The restoring torque produced by the weight is therefore 68.6 newton-meter(N-m). The person is on the verge of toppling when the magnitudes of thesetwo torques are just equal; that is, T, =T, ornicPostureWictioning at anFa × 1.5m=68.6N – m(1.4)on at the HipTherefore, the force required to topple an erect person is68.6Fa=1.5e Fin of a45.7N (10.31b)(1.5)cribology. OralribologyActually, a person can withstand a much greater sideways force withoutlosing balance by bending the torso in the direction opposite to the applied2. Haircribology3. BiotribologyTextilesciseser 3: Translational78°FENGWre to searchTO

Given, Fa=45.7N Let the friction force be on the body fs and the friction coefficient is μs. To…

Answered: 87 F, 1.5 m A Pivot point 0.1 m FIGURE…

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