A lunch tray is being held in one hand, as the figure below illustrates.
The mass of the tray itself is 0.182 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.08 kg plate of food and a 0.279 kg cup of coffee. Assume L1 = 0.0590 m, L2 = 0.138 m, L3 = 0.258 m, L4 = 0.381 m and L5 = 0.401 m. Obtain the force T exerted by the thumb. This force acts perpendicular to the tray, which is being held parallel to the ground.
To solve this problem, we need to analyze the torques acting on the lunch tray and set up an equation using the principle of torque equilibrium.
Torque is calculated by multiplying the force applied to an object by the distance from the applied force to the pivot point. In this case, the pivot point is the center of gravity of the tray.
Let's label the forces acting on the tray:
- T: The force exerted by the thumb.
- Weight of the tray: W1 = m1 * g (m1 is the mass of the tray and g is the acceleration due to gravity).
- Weight of the plate of food: W2 = m2 * g (m2 is the mass of the plate of food).
- Weight of the cup of coffee: W3 = m3 * g (m3 is the mass of the cup of coffee).
We can now calculate the torques produced by each force:
- Torque from force T: τ1 = T * L1 (since the force is perpendicular to the tray).
- Torque from the weight of the tray: τ2 = W1 * L2 (since the weight acts downwards at the center of gravity of the tray).
- Torque from the weight of the plate of food: τ3 = W2 * L3.
- Torque from the weight of the cup of coffee: τ4 = W3 * L4.
According to the principle of torque equilibrium, the sum of all torques acting on an object must be zero. Therefore, we can set up the equation:
τ1 + τ2 + τ3 + τ4 = 0
Substituting the expressions for the torques:
T * L1 + W1 * L2 + W2 * L3 + W3 * L4 = 0
Now, we can substitute the expressions for the weights of the tray, plate of food, and cup of coffee:
T * L1 + (m1 * g) * L2 + (m2 * g) * L3 + (m3 * g) * L4 = 0
Finally, we can solve for the force T:
T = -((m1 * g) * L2 + (m2 * g) * L3 + (m3 * g) * L4) / L1
Substituting the given values and solving the equation will give us the force T exerted by the thumb.
To obtain the force T exerted by the thumb, we need to consider the rotational equilibrium of the lunch tray system.
Step 1: Calculate the torque exerted by the tray's weight.
- The weight of the tray itself exerts a torque about the pivot point created by the thumb. The torque is given by the formula:
Torque1 = (mass of the tray) × (gravitational acceleration) × (distance from pivot)
= (0.182 kg) × (9.8 m/s^2) × (L2 + L3 + L4 + L5)
= 32.38 N·m
Step 2: Calculate the torque exerted by the plate of food.
- Since the tray is being held parallel to the ground, the plate of food does not exert any torque.
Step 3: Calculate the torque exerted by the cup of coffee.
- The cup of coffee exerts a torque about the pivot point. The torque is given by the formula:
Torque3 = (mass of the cup) × (gravitational acceleration) × (distance from pivot)
= (0.279 kg) × (9.8 m/s^2) × (L2 + L3 + L4 + L5 - L1)
= 3.23 N·m
Step 4: Calculate the force T exerted by the thumb.
- Since the system is in rotational equilibrium, the total torque exerted by the system must be zero.
- Therefore, the force T exerted by the thumb must be equal in magnitude and opposite in direction to the total torque exerted by the tray's weight and the cup's weight.
- This can be expressed as:
Torque1 + Torque3 = T × L1
32.38 N·m + 3.23 N·m = T × 0.0590 m
35.61 N·m = T × 0.0590 m
T ≈ 604.91 N
Therefore, the force T exerted by the thumb is approximately 604.91 N, and it acts perpendicular to the tray.