A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.280 kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00 kg plate of food and a 0.210 kg cup of coffee. Obtain the force T exerted by the thumb and the force F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground.
T = N (downward)
F = N (upward)
W1= weight of the tray 0.280 kg
W2= weight of the plate 1.0 kg
W3= weight of the cup 0.210 kg
X1= center of gravity of tray is at geographical center 0.200m
X2= center of gravity of plate is at geographical center with respect to its position on tray 0.240m
X3= center of gravity of cup is at geographical center with respect to its position on tray 0.380m
Xcg= (W1X1)+(W2X2)+(W3X3)/ W1+W2+W3
Xcg= [(0.280kg)(0.200m)]+[(1.0kg)(0.240m)]+[(0.210kg)(0.380m)]/(0.280+1.0+0.210)
Xcg= (0.056+0.240+0.0798)/1.49
Xcg= 0.3758/1.49
Xcg= 0.2522m
I have found the center of gravity for the whole system which is 0.2522m. How to I proceed to answer the question?
I don't know why you did that.
The solution is to sum the moments about the edge (or any point) and set to zero. From the description, I am not certain if the thumb or the fingers are on the edge.
You have that equation, and one other: the sum of the vertical forces is zero.
That will be enough to solve.
The thumb is on top of the tray extending 0.0600m from the edge of the tray inward and exerting T downward.
The other four finger and beneath the tray entending 0.100m from the edge of the tray inward and exeting a force F upwards.
I still don't understand how to tackle this question. I thought you had to fing the center of gravity for the tray, plate and cup then using that find T and F. Please help!!!!
I apologize for the confusion. Let's start by summing the moments about the edge of the tray to find the forces T and F.
To do this, we need to consider the torques (moments) exerted by each force. The torque of a force is calculated by multiplying the force by the perpendicular distance from the axis of rotation.
In this case, the axis of rotation is the edge of the tray. The thumb exerts a force T downward at a distance of 0.0600m from the edge, and the fingers exert a force F upward at a distance of 0.100m from the edge.
The torque equation about the edge of the tray is:
(T × 0.0600m) + (F × 0.100m) - (W1 × X1) - (W2 × X2) - (W3 × X3) = 0
Now, let's substitute in the known values:
(T × 0.0600m) + (F × 0.100m) - (0.280kg × 0.200m) - (1.0kg × 0.240m) - (0.210kg × 0.380m) = 0
Simplifying the equation, we have:
0.0600T + 0.100F - 0.0560 - 0.240 - 0.0798 = 0
0.0600T + 0.100F = 0.3758
This is our first equation.
Now, let's consider the second equation by summing the vertical forces. Since the tray is being held parallel to the ground, the sum of the vertical forces is zero.
The vertical forces are T (downward), F (upward), and the weights of the tray, plate, and cup (all downward).
T + W1 + W2 + W3 = 0
Substituting the known values:
T + 0.280kg + 1.0kg + 0.210kg = 0
T + 1.49kg = 0
T = -1.49kg
This is our second equation.
Now, we have a system of equations:
0.0600T + 0.100F = 0.3758
T = -1.49kg
We can use these equations to solve for the force F exerted by the four fingers.
Substituting the value of T into the first equation:
0.0600(-1.49kg) + 0.100F = 0.3758
-0.0894kg + 0.100F = 0.3758
0.100F = 0.3758 + 0.0894kg
0.100F = 0.4652kg
F = 0.4652kg / 0.100
F ≈ 4.652N
Therefore, the force F exerted by the four fingers is approximately 4.652N.
Please note that the negative sign on T indicates that it is directed downward, as mentioned in the problem statement.
I hope this clears up any confusion. Let me know if you have any further questions!