A Lunch Tray Due in 11 hours, 42 minutes A lunch tray is being held in one hand,

Question

A Lunch Tray Due in 11 hours, 42 minutes

A lunch tray is being held in one hand, as the figure below illustrates.

The mass of the tray itself is 0.189 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.10 kg plate of food and a 0.251 kg cup of coffee. Assume L1 = 0.0600 m, L2 = 0.121 m, L3 = 0.242 m, L4 = 0.383 m and L5 = 0.405 m. Obtain the magnitude of the force T exerted by the thumb. This force acts perpendicular to the tray, which is being held parallel to the ground.

A Lunch Tray Due in 11 hours, 42 minutes A lunch tray is being held in one hand, as the figure below illustrates. The mass of the tray itself is 0.189 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.10 kg plate of food and a 0.251 kg cup of coffee. Assume L1 = 0.0600 m, L2 = 0.121 m, L3 = 0.242 m, L4 = 0.383 m and L5 = 0.405 m. Obtain the magnitude of the force T exerted by the thumb. This force acts perpendicular to the tray, which is being held parallel to the ground.

Explanation / Answer

By the equilibrium of forces

F=T+(0.189+.251+1.10)*9.8 =>F=T+15.092 -------------------------------------------------------(1)

By net torque =0 about the hinge from the hand

T*L1+1.1*9.8*L3+0.251*9.8*L4+.189*9.8*L5/2=F*L2 -------------------------------------------(2)

0.06T+2.608+0.9132+0.375=(T+15.092)0.121

0.06T+3.8962=0.121T+1.82

.061T=2.06

T=33.77 Newton

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