Respuesta :
(a)The value of W for the given condition will be 237.3 N.
(b)The distance must the fulcrum be moved to restore balance will be 0.070 m.
What is the center of mass?
A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.
Given data;
Weight is applied to the right end of the rod, W₁= 225 N
Rod's weight,W₂ = 255
W₃ is the Weight at "x" distance from the left end.
x1 is the location of W₁ with respect to the left end, where L = 1.80 m
m is the location of the rod's center of mass from the left end.
The position of W from the left end; x₃ = 50 cm = 0.50 m
Xcm is the position of the center of gravity of the rod at the fulcrum from the left end;
Given that the fulcrum is balanced in this situation, the rod's center of gravity will be there.
Xcm= L - 75 cm
Xcm=1.80 m - 0.75 m
Xcm= 1.05 m
The value of x₂
x₂= 1.80 m/2 = 0.90 m
Currently, the horizontal center of gravity is determined by
[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]
Substitute the values we get;
[tex]\rm 1.05 = \fraca{ (225 \times 1.80 + 255\times 0.90 + W\times 0.50)}{(225 + 255 + W)}\\\\ W = \frac{ (225\times 1.80 + 255\times 0.90 - 480 \times 1.05)}{(1.05 - 0.50)}\\\\ W = 237.3 \ N[/tex]
After moving the fulcrum 20.0 cm to the right,
The location of the W from the left end;
x₃ = .50 cm + 20 cm
x₃ = 70 cm
x₃ = 0.70 m
W₁, W₂, W, x₁, and x₂ have fixed values. Therefore, based on these values, the new center of gravity will be:
[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3*x_3)}{(W_1 + W_2 + W_3)}[/tex]
[tex]\rm X_{cm} = \frac{(225\times 1.80 + 255\times 0.90 + 237.3\times 0.70)}{(225 + 255 + 237.3)}\\\\ X_{cm} = 1.12 m[/tex]
Fulcrum must be shifted by;
Y = (1.12 - 1.05)
Y= 0.070 meters.
W has been moved in the appropriate direction, thus the fulcrum should likewise be moved in the appropriate direction.
Hence, the value of W for the given condition will be 237.3 N. and the distance must the fulcrum be moved to restore balance will be 0.070 m.
To learn more about the center of mass refer;
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