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.
I've tried this and I don't understand anybody's work
To obtain the force T exerted by the thumb on the tray, we can use the principle of moments or torque. Torque is the product of force and the perpendicular distance from the point of rotation. In this case, the point of rotation is the center of gravity of the tray.
To find the torque exerted by the thumb, we consider the torques produced by different objects on the tray.
1. Torque produced by the tray's own weight:
The weight of the tray acts vertically downward through the center of gravity. Since the center of gravity is located at the geometrical center of the tray, the torque produced by the tray's weight is zero.
2. Torque produced by the plate of food:
The weight of the plate of food can be considered acting on its center of gravity, which is also the center of the tray. Since the distance from the center of the tray to itself is zero, the torque produced by the plate of food is also zero.
3. Torque produced by the cup of coffee:
Similar to the plate, the weight of the cup of coffee acts on its center of gravity. Since the distance from the center of the tray to itself is zero, the torque produced by the cup of coffee is also zero.
Therefore, the only torque remaining is the one produced by the force T exerted by the thumb.
The torque exerted by the thumb can be calculated using the equation:
Torque = Force * Perpendicular Distance
In this case, the perpendicular distance is given by L5 since the force acts perpendicular to the tray. The values are given as L5 = 0.401 m.
So, we have:
Torque = T * L5
Since there are no other torques, this torque should be balanced by the torques produced by the weights of the tray, plate, and cup.
Considering the weights and their distances from the center of the tray, we can write the equation:
Torque by thumb = Torque by plate + Torque by cup + Torque by tray
T * L5 = (Plate weight) * L2 + (Cup weight) * L3 + (Tray weight) * 0
Now, let's substitute the given values and solve for T:
T * 0.401 = (1.08 kg) * 0.138 m + (0.279 kg) * 0.258 m + (0.182 kg) * 0
T * 0.401 = 0.14904 kg⋅m + 0.071682 kg⋅m + 0
T * 0.401 = 0.220722 kg⋅m
T = 0.220722 kg⋅m / 0.401
T ≈ 0.550 N
Therefore, the force exerted by the thumb on the tray is approximately 0.550 Newtons.