Question

A nonuniform, horizontal bar of mass is supported by two massless wires against gravity. The left wiremakes an angle $phi_1$ with the horizontal, and the right wire makes an angle $phi_2$ . The bar has length .

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Find the position of the center of mass ofthe bar, ,measured from the bar's left end.

Express the center of mass in termsof , $phi_1$ , and $phi_2$ .

Answer 1

Concepts and reason

The concepts required to solve the problem are the tension, rotational and translational equilibrium force, and center of mass.

Consider the translational equilibrium force for the block to determine the tension on the left rope. Use the tension in the rotational equilibrium for the block. Rearrange the equation of rotational equilibrium to calculate the center of mass.

Fundamentals

The tension on the rope will have both vertical and horizontal components.

At the translational equilibrium, the forces acting vertically and horizontally will be balanced. The net force acting horizontally and the net force acting vertically will be equal to zero to balance the block without breaking the rope.

At the rotational equilibrium, the torques acting vertically and horizontally will be balanced. The net torque acting horizontally and the net force acting vertically will be equal to zero to balance the block without breaking the rope. The torque is the product of the force and perpendicular distance to which force is acting.

The block is hanged as shown in figure 1.

Figure 1

The translational equilibrium of the block along vertical axis is,

Here, is the tension on the left rope, is the angle made by the left rope with the horizontal axis, is the tension on the right rope, is the angle made by the right rope with the horizontal axis, is the mass of the block and is the acceleration due to gravity.

The translational equilibrium of the block along horizontal axis is,

Rearrange the equation in terms of the tension on the left rope. The tension on the left rope is,

The rotational equilibrium of the block is,

Here, is the center of mass and is the length of the block.

Substitute for in the rotational equilibrium of the block. The rotational equilibrium of the block is,

.

Rearrange the equation in terms of the center of mass to calculate the center of mass of the block. The center of mass of the block is,

The expression for center of mass is .

Ans:

The expression for center of mass is .