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3 votes
In the top-view diagram a special case is shown as an

example. One mass is set at the 0° location, another
at the 90° location, and the third at the 196⁰.
Perspective View
-
180°
190°
270°
Top View
Given the first two masses values and locations:
m1
461 g, 0₁ = 0.00°, m₂ = 476 g, and
01
-
=
02 100°. Find the angle 03 and the mass m3 that
can be placed so that the object remains at rest.

User Alchemy
by
8.2k points

1 Answer

5 votes

To solve this problem, we can use the following steps:

Draw a free-body diagram for each mass. This will help us to identify all of the forces acting on each mass.

Write down the equations for static equilibrium. This means that the net force and net torque on each mass must be zero.

Solve the equations for the unknown angle 03 and the mass m3.

Here is a more detailed explanation of each step:

Draw a free-body diagram for each mass.

The free-body diagram for each mass will look like this:

m1

/ \

/ \

F_g T_1

\ /

\

m2

/ \

/ \

F_g T_2

Where:

m1 and m2 are the masses of the first two objects, respectively.

F_g is the weight of each object.

T_1 and T_2 are the tensions in the strings connecting the objects.

Write down the equations for static equilibrium.

The equations for static equilibrium are:

Net force = 0

Net torque = 0

For the first object, the equations for static equilibrium are:

F_g - T_1 = 0

T_1 * d1 = T_2 * d2

Where d1 and d2 are the distances from the pivot point to the first and second objects, respectively.

For the second object, the equations for static equilibrium are:

F_g - T_2 = 0

T_2 * d2 = T_3 * d3

Where d3 is the distance from the pivot point to the third object.

Solve the equations for the unknown angle 03 and the mass m3.

To solve for the unknown angle 03, we can use the following equation:

tan(03) = (T_2 * d2) / (T_1 * d1)

To solve for the unknown mass m3, we can use the following equation:

m3 = (T_2 * d2) / g

Where g is the acceleration due to gravity.

Substituting the known values into the equations:

m1 = 461 g, d1 = 0.00 m, m2 = 476 g, d2 = 1.00 m, and 02 = 100°

We can now solve for 03 and m3.

Solving for 03:

tan(03) = (476 g * 1.00 m) / (461 g * 0.00 m)

tan(03) = ∞

03 = 90°

Solving for m3:

m3 = (476 g * 1.00 m) / 9.81 m/s^2

m3 = 49 g

Therefore, the angle 03 is 90° and the mass m3 is 49 g.

Checking our answer:

We can check our answer by substituting the values of 03 and m3 into the equations for static equilibrium.

For the first object:

F_g - T_1 = 0

m1 * g - T_1 = 0

461 g * 9.81 m/s^2 - T_1 = 0

T_1 = 461 g * 9.81 m/s^2

T_1 * d1 = T_2 * d2

461 g * 9.81 m/s^2 * 0.00 m = T_2 * 1.00 m

461 g * 9.81 m/s^2 = T_2

For the second object:

F_g - T_2 = 0

m2 * g - T_2 = 0

476 g * 9.81 m/s^2 - T_2 = 0

T_2 = 476 g * 9.81 m/s^2

T_2 * d2 = T_3 * d3

476 g * 9.81 m/s^2 * 1.00 m = T_3 * d3

T_3 = 476 g

User Martinlasek
by
9.1k points