Question

Find the amount of torque the “thumb” (the black dot) exerts on the tray of food in order to reach equilibrium as well as the total amount of force.

4.5 cm dky Msoda 450 kg 250 ТА 75cm

Answer 1

Forces acting on the tray.

1 kg = 9.8 N at 4.5 cm

0.250 kg = 0.250 *9.8 N at 75/2 cm , weight of the tray

0.450 kg = 0.450*9.8 N at 75 cm (soda) end of the tray

all these force produce CW torque, forces acting down ward

Total torque

= 9.8* 0.045 + 0.250*9.8 * 75/200 + 0.450*9.8 *75/100

= 4.67 N-m

Thumb has to produce equal torque CCW for equilibrium.

Thumb torque = 4.67 N-m

Thumb force F_th - torque arm = 3.5 cm

F_th * 0.035 = 4.67 N-m

F_th = 133.4 N