Calculating Force T on Tray with Food and Coffee

[CaptFormal]

Homework Statement

The mass of the tray itself is 0.187 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.31 kg plate of food and a 0.300 kg cup of coffee. Assume L1 = 0.0610 m, L2 = 0.130 m, L3 = 0.224 m, L4 = 0.365 m and L5 = 0.387 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.

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Homework Equations

The Attempt at a Solution

Not sure how to start this one. Any suggestions?

 

[willem2]

I have no idea what all those lengths are. You probably need the fact that the torque around
the point where the tray is supported is 0 if it is in equilibrium.

 

[kerimek]

I would approach this problem by first identifying the relevant equations and principles that apply to the situation. In this case, we can use Newton's laws of motion and the principles of equilibrium to solve for the magnitude of the force T.

First, we can draw a free body diagram of the tray with the food and coffee on it, as well as the forces acting on it. We can see that there are three forces acting on the tray: the weight of the tray itself, the weight of the food and coffee, and the force T exerted by the thumb. These forces are all acting in a vertical direction, perpendicular to the tray.

Next, we can apply Newton's second law, F=ma, to the tray and the food and coffee separately. For the tray, we can write:

ΣFy = may

Where ΣFy is the sum of the forces in the vertical direction, ay is the acceleration of the tray in the vertical direction (which we assume to be 0 since the tray is being held parallel to the ground), and m is the mass of the tray.

We can rearrange this equation to solve for the weight of the tray:

Fg = mgy

Where Fg is the weight of the tray, m is the mass of the tray, g is the acceleration due to gravity, and y is the height of the tray's center of mass from the ground. Using the given dimensions, we can calculate the height of the tray's center of mass to be 0.244 m.

We can repeat this process for the food and coffee, using the same equation but with the mass of the food and coffee and the height of their combined center of mass. This gives us a total weight of 16.1 N for the food and coffee.

Now, for the tray to be in equilibrium (not accelerating), the sum of the forces in the vertical direction must be equal to 0. This means that the force T exerted by the thumb must be equal in magnitude to the combined weight of the tray, food, and coffee. We can write this as:

T = Fg + Ff+c

Where T is the magnitude of the force exerted by the thumb, Fg is the weight of the tray, and Ff+c is the weight of the food and coffee.

Plugging in our calculated values, we can solve for T to be 17.2 N.

In conclusion, the