Question

xiv) A, 2 m long wooden plank of mass 20
kg is pivoted (supported from below) at
0.5 m from either end. A person of mass
40 kg starts walking from one of these
pivots to the farther end. How far can the
person walk before the plank topples?​

Answer 1

FOR DIAGRAM SEE THE ATTACHMENT

Now from the laws of equilibrium

∑F=0⇒R1−40g−45g+R2=0⇒R1+R2=85g=850 N

and

∑τ=0⇒R1×0.5−40g×2.5−45g×3.5+R2×4.5=0⇒R1+9R2=5150

Solving for R1 and R2

R1=312.5 NR2=537.5 N

HOPE THIS WILL HELP. PLEASE MARK AS BRAINLIEST AND FOLLOW ME.

Answer 2

Answer:

Explanation:

Before the cable breaks, all of the torques and forces are in equilibrium. Draw a free body

diagram and use it to write net force and net torque equations. Measuring torques about the

given pivot will eliminate the unknown force from the pivot. You’ll be left with torques from

the weight of the plank, weight of the man, and the tension. Using the maximum value of the

tension as the measured value of the tensional force, you can find the maximum distance that the

man can walk and still have his torque balanced so that the cable does not break. You will find

that he can walk 0.57 m.