Equilibrium of a rigid body lab report. Solved: Torques And Rotational Equilibrium Of A Rigid Body... 2019-02-16

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Physics 4A zhong: Lab 11: Balanced Torques and Center of Gravity

equilibrium of a rigid body lab report

Let the unknown point equal to x cm. We are carrying a huge 3D Digital Library ready to use. The lever arm is the perpendicular distance from O to the line of action of the force. Record the masses and positions. The three forces acting on a body which is in equilibrium may be either concurrent or parallel. The body will be in equilibrium if the resultant force in horizontal direction is zero i.

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Equilibrium of a Rigid Body

equilibrium of a rigid body lab report

Compare the calculated and experimental result. Is it necessary to include the mass of the clamps in your caculation? Replace one of the above masses with an unknown mass. . Calculate the position of the center of gravity of this combination two masses and meter stick. A non-concurrent force systems will be in equilibrium if the resultant of all forces and moment is zero.

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Equilibrium of a Rigid Body STF3

equilibrium of a rigid body lab report

If the object is first balanced to find its center of mass, then the entire weight of the object can be considered to act at that center of mass. A rigid body is said to be in rotational equilibrium, if the body does not rotate or rotates with constant angular velocity. Place the same two masses used above at different locations on the same side of the support and balance the system with a third mass on the opposite side. The three forces are acting in opposite direction and their magnitude is so adjusted that there is no resultant force and body is in equilibrium. Hence a body will not be in equilibrium under the action of two equal and opposite parallel forces. Principle Of Moments:According to the principle of moments, a body will be in rotational equilibrium if algebraic sum of the moments of all forces acting on the body about a fixed point is zero. When a body is subjected to two forces, then the body will be in equilibrium if the two forces Fig.


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Solved: Torques And Rotational Equilibrium Of A Rigid Body...

equilibrium of a rigid body lab report

The center of gravity of the body coincides with the center of mass of the body. The resultant moment about point A is given by ยท Fig. With the 200 grams still at 90 cm mark, imagine that you now position an additional 100 grams mass at the 30 cm mark on the meter stick. This problem can be solved analytically and graphically. Suspend the body from some point such as A.

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Equilibrium of a Rigid Body

equilibrium of a rigid body lab report

Thus for concurrent force system, the condition I. Some masses have been rusted, some their masses are not accurate. The five forces F 1, F 2, F 3, F 4 and F 5are acting at a point on a body as shown in Fig. If the two forces acting on a body are equal and opposite but are parallel, as shown in Fig. We learned that t orque is the tendency of a force to rotate an object about an axis, fulcrum, or pivot. Place about 200 grams at 90 cm on the meter stick and balance the system by changing the balance point of the meter stick. As the body is in equilibrium, the resultant forcein x-direction and y-direction should be zero.

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Physics 4A zhong: Lab 11: Balanced Torques and Center of Gravity

equilibrium of a rigid body lab report

Purpose: To investigate the conditions for rotational equilibrium of a rigid bar and to determine the center of gravity of a system of masses. We also customise the content as per your requirement for companies platform providers colleges etc. If the magnitude of the force F1 is 100 N and its acting at 0 long x-axis as shown in Fig. From this information, calculate the mass of the meter stick. Should the clamp holding the meter stick be included as part of the mass of the meter stick? Hence for three concurrent forces acting on a body when the body is in equilibrium, the resultant of the two forces should be equal and opposite to the third force. For the equilibrium of the body, the resultant force in the vertical direction should be zero here there is no force in horizontal direction. For this, the net external torque or the vector sum of all the torques acting on the body is zero.

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Solved: Torques And Rotational Equilibrium Of A Rigid Body...

equilibrium of a rigid body lab report

So, here we still use the balance point as 48. Meter stick Data Table Data and Calculations Table 1 Data and calculations Table 2 Data and calculations table 3. Sum the torques about your pivot point O and compare with the expected value. This sign represents the algebraic sun of forces or moments. This is shown in Fig.

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Physics 4A zhong: Lab 11: Balanced Torques and Center of Gravity

equilibrium of a rigid body lab report

Let us suppose that F2 is acting in opposite direction as shown in Fig. The resultant of F 1 and F 2 is given by R. A body is in rotational equilibrium, when algebraic sum of moments of all the forces acting on the body about a fixed point is zero. Substituting the value of e in equation i , we get F 5cos 63. When two forces are acting on a body and the body is in equilibrium, then the two forces should be collinear, equal and opposite. Calculate the net torque on this system about the point support and compare with the expected value.

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