Examples static of rigid bodies

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Example 1

Calculate the resultant, at point A, of the forces applied to the piece.

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Points, four elements Point, one element Resultant and three elements Force, as show the image below:

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To the element Points, we call this Placa (Piece):

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To the element Point that represent the corner A, assuming this is at the origin, we have:

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Now to the element Point at corner B:

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And to the applied force at this corner:

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To the element Point at corner C:

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And to the applied force here:

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To corner D:

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And to the force:

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Now to the element Resultant:

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Now click in the icon Solve to get the answer:

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Example 2

Calculate the resultant, at point O, of the applied force (ignore the weight of the beam).

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the English system. And add one element Stationary reference system, one element Beam, two elements of beam, one element Force and one element Resultant with vertical force, as show the image below:

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The element Stationary reference system set, to default, the gravity value:

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To the element Beam, we call this Barra (Beam), we have:

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Here we have 0 to the mass, because we must ignore the weight of the beam. To the element of beam that represent the lower side, we have:

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And to the element Resultant applied here:

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To the other element of beam, we have:

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And to the applied force:

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Now click in the icon Solve to get the answer:

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Example 3

If we want to replace the two forces of 175 N and 380 N force with a single force applied at point P as shown the image, What should be the distance from point P to point O and the magnitude of this force? (The beam and the drum form a rigid body)

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. This problem can be solved on different ways, here we will solve it as follow. First add one element Points, three elements point and the necessary elements force and resultant, as show the image below. Our initial intention is calculate the couple of the pair of 175 N forces, at point O. 

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To the element Points, that represent the different points of the drum and that we call Tambor (drum), we have:

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To the element point that represent the center O, we assume this at origin:

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And to the element Resultant applied here:

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Now to the upper point, we have:

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And to the lower point:

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And to the two applied forces:

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Now click in the icon Solve to get the answer:

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Now that we know the moment of the pair of 175 N forces, erase the content of chalkboard and add the necessary elements as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Beam, we call this Viga (beam), we have:

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Here we ignore the weight of the beam. To the element of beam that represent the lower side and to the applied couple (the previous calculated value), we have:

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Now to the element of beam where is applied the other force, and to this force, we have:

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To the element of beam that represent the point P:

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Here the distance is an unknown data. And to the resultant applied here:

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Here the couple is 0, because we want just a force, which is an unknown data. Now click in the icon Solve to get the answer:

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Example 4

Calculate the resultant, at joint O, of the applied forces.

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Beam, three elements of beam and the necessary elements Force and Resultant, as show the image below:

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The gravity value is added to default. Now to the element Beam, that we call Viga (beam), we have:

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To the element of beam that represent the lower side, and to the resultant applied here, we have:

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On the resultant element all are unknown data. Now to the element of beam that represent the point A, and to the applied force, we have:

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To the element of beam that represent the point B and to the applied force, we have:

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Now click in the icon Solve to get the answer:

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As we expected the resultant is only a couple. Here the angle is irrelevant.

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Example 5

Calculate the support reactions, knowing that the beam has a mass of 4 kg and a length of 2.1 m.

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Beam, six elements of beam and the necessary elements Force, as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Beam, that we call Viga (beam) and assuming has its center of mass at middle, we have:

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Now to the elements of beam and its applied forces we have, from left to right:

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To the element at 40 cm from the left side, we call RA the reaction:

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At 85 cm from the left side:

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At 110 cm:

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At 160 cm, we call RB the reaction:

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And at 195 cm, the last element:

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Now click in the icon Solve to get the answer:

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Example 6

If the tension in the cable BC is 155 N, what are the reactions at A? (The mass of the beam is 3 kg and its center of mass is 75 cm from A)

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Beam, three elements of beam and the necessary elements Force and Couple, as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Beam, that we call Viga (beam):

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To the element of beam at side A, we have:

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And to the forces and couple applied, all unknown data, we have:

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To the element of beam where the mass is hanging:

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And to the applied force, the weight of the mass (18 kg*9.81 m/s2 = 176.58 N), we have:

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And to the element of beam that represent the right side:

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And to the applied force by the cable:

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Now click in the icon Solve to get the answer:

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Example 7

What is the maximum force F that can be applied without tipping the piece? (The mass of the piece is 3.5 kg and the mass center point is indicated by the 18 cm).

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Solid, two elements of solid and the necessary elements Force, as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Solid, that we call Pieza (piece), we have:

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To the element of solid that represent the corner where is applied the force F, we have:

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And to the applied force here, that is an unknown data:

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The other element of solid represent the lower right corner, because when the tipping is imminent, the normal force is applied here. Therefore, we have:

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And to the applied force, that represent the normal force and that is an unknown data:

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Now click in the icon Solve to get the answer:

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Example 8

A uniform block of 4 kg, as show the image, is placed on a slope. What is the maximum angle of inclination of the plane, so that the block does not slide or tipping (whichever comes first)? (The coefficient of friction between the block and the plane is 0.4).

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Solid, one element of Solid, one element Angles and the necessary elements Force and Friction, as show the image below:

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Our intention to introduce a frictional force is find the angle when the slip is imminent. Thus, the block element represents the point where is applied the normal and here we apply the friction force too (we can move the friction along its action line, so we can apply it in this element).

The Stationary reference system set the gravity value to default. Now on the element Solid, that we call Bloque (block), we have:

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Here the angle is an unknown data. To the element of Solid, we have:

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The x coordinate is unknown. This must be so. Because when the sliding of block is imminent, we don’t know at where is applied the normal force. Now to the normal force:

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To the element Friction:

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The angle is the same as of the block. Because the friction is parallel to the plane. Now to the element Angles, we have:

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Because these angles must be complementary. Once entered the data, click on the icon Solve to get the answer:

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Then, the angle when the sliding of block is imminent is 21.801° (As we expected, because arctan 0.4 = 21.801). In the case of a negative value, this would mean that the block has already overturned.

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Example 9

A uniform ladder, of 3.5 kg and 3.8 m long, rests against a wall as show the image. If a child with a mass of 40 kg, ups to the fourth step at 3.2 m from the base, find the angle for which this position corresponds to the slip imminent if the coefficient of friction between the ladder and the floor is 0.35 (there is no friction between the ladder and the wall).

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Beam, three elements of beam and the necessary elements Force and Friction, as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Beam, that we call Escalera (ladder), recalling that the ladder is uniform and that the angle is unknown:

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To the element of beam that represent the lower side, we have:

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And to the normal force and the friction:

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Now to the element that represent the point where the child is supported, we have:

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And to the force that represent the weight of this child (40 kg*9.81 m/s2 = 392.4 N):

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Finally, to the element that represent the upper side:

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And to the force that represent the reaction of the wall:

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Now click in the icon Solve to get the answer:

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Example 10

Consider the structure in the image below. Calculate reactions at O and reactions between the beams at points A, B and C. Ignore the weight of the beams.

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, three elements Beam, seven elements of beam and the necessary elements Force and Couple, as show the image below (the yellow lines separate the areas where are defined the beams):

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In the image can be seen that, according to Newton’s third law, the reactions between the beams have a different sense. And we assign the same magnitude or unknown data. The Stationary reference system set the gravity value to default. Now to the element Beam that represent the vertical beam, that we call OAB:

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This beam is composed of three elements, which represent the points O, A and B. For the element in O, and their respective reactions, called Rx, Ry, M, we have:

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For the element in A and their respective reactions, RAx and RAy, we have:

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For the element in B and their respective reactions, RBx and RBy, we have:

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Now for the element Beam representing the horizontal beam, that we call AC, we have:

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For the element of beam that represent the point A, and to its respective reactions, we have:

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To the element of beam at point C, for the reactions RCx, RCy and to the weight of 50 kg (50 kg*9.81 m/s2 = 490.5 N), we have:

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Now to the element Beam that represent the beam from B to C, that we call BC (here the angle is entered as slope -0.7/1.3):

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To the element of beam at B and to its reactions, we have:

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For the element of beam at C, here we calculate the distance between B and C, which is approximately 1.48 m (see note at the end of this example on how to avoid this problem), we have:

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And to the reactions RCx and RCy (we don’t add here the weight of the 50 kg, because was added at the beam AC):

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Now click in the icon Solve to get the answer:

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Note: A more easy way to solve this problem (that avoid having to calculating the length of the beam BC), is replace the BC beam with two elements Beam of 2 forces (notice that the beam BC is an element that supports two forces). See image below. Everything is a matter of deciding which is best for the problem to solve. The beam AC is also a beam of two forces. However, we can’t replace both.

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Example 11

A uniform block of 25 kg and 50 cm of edge, as show the image below, is placed on a slope of 20°. If we want move up the block as fast as possible (constant speed) by the horizontal force F, What is the maximum possible coefficient of friction? Which is the force?

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the SI system. And add one element Stationary reference system, one element Solid, two elements of solid and the necessary elements Force and Friction, as show the image below:

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The Stationary reference system set the gravity value to default. Now to the element Solid, that we call Bloque (block) we have:

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To the element of solid that represent the upper corner, where is applied the force F, we have:

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And to the horizontal force:

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The normal force and the friction are applied to the element of solid that represent the lower right corner. Because when we move up the block as fast as possible, the tipping is imminent. Then, we have:

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To the force that represent the normal force:

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And to the friction:

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Now click in the icon Solve to get the answer:

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Example 12

In the structure of the image all the joints are pin joints. Ignoring the weight of the beams, get all the reactions in the joints.

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Solution with FisicaLab

Notice that the beams BE and CF are beams of two forces. So, we build the problem with two elements Beam, and the necessary elements Force and Beams of 2 forces, as show below:

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The Stationary reference system set the gravity value to default. Now to the element Beam that represent the ABC beam, we have:

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This beam have three elements: A, B and C. To the element that correspond with A:

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And to the applied Forces, the reactions:

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Now to the element that correspond with B:

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And to the element Beam of 2 forces, that correspond with the beam from B to E, the angle is entered as the slope 22/50:

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To the element that correspond with C:

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To the element Force, that correspond with the force of 450 N:

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And to the element Beam of 2 forces, that correspond with the beam from C to F, the angle is entered as the slope 23/50:

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Now to the element Beam that represent the DEF beam, we have:

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This beam have three elements: D, E and F. To the element that correspond with D:

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And to the applied Forces, the reactions:

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To the element that correspond with E:

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And to the element Beam of 2 forces, that correspond with the beam from B to E, the angle is also the slope 22/50:

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To the element that correspond with F:

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And to the element Beam of 2 forces, that correspond with the beam from C to F, the angle is also the slope 23/50:

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Now click in the icon Solve to get the answer:

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Example 13

In the truss of the image, find the support reactions and internal forces in all beams under the load conditions shown.

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Solution with FisicaLab

Select the group Static and, inside this, the module Static of solids. Erase the content of the chalkboard and select the English system. Then add one element Truss, eight elements Joint of truss, eight elements Force, five forces for the load forces and three for the support reactions (notice that in support A there is only one vertical force), and 26 elements Beam of truss to build the problem as show below:

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To the element Truss, we use the name AH:

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All the elements Joint of truss, will owned by this truss. So, we add each joint to this truss:

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For the two elements Beam of truss that represent the beam from joint A to C, and calling AC the internal force, we have:

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Now for the two elements Beam of truss that represent the beam from joint A to B, and calling AB the internal force and entering the angle as the slope 6/8, we have:

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For the two elements Beam of truss that represent the beam from joint B to C, and calling BC the internal force, we have:

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For the two elements Beam of truss that represent the beam from joint B to E, and calling BE the internal force, we have:

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Here the angle is entered as slope 28/96. The numerator and denominator should be at the same units: 2 ft 4 in = 28 in and 6 ft = 96 in. Also can be used 2.333/6 using the feet unit. Now for the two elements Beam of truss that represent the beam from joint B to E, and calling BE the internal force, we have:

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Here the angle is also the slope 6/8. For the two elements Beam of truss that represent the beam from joint C to E, and calling CE the internal force, we have:

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For the two elements Beam of truss that represent the beam from joint D to E, and calling DE the internal force, we have:

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Now for the two elements Beam of truss that represent the beam from joint D to F, and calling DF the internal force, we have:

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Here the angle is entered as slope 28/96. For the two elements Beam of truss that represent the beam from joint E to F, and calling EF the internal force, we have:

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Here the angle is entered as slope 6/8. Now for the two elements Beam of truss that represent the beam from joint E to G, and calling EG the internal force, we have:

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For the two elements Beam of truss that represent the beam from joint F to G, and calling FG the internal force, we have:

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For the two elements Beam of truss that represent the beam from joint F to H, and calling FH the internal force, we have:

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Here the angle is entered as slope 6/8. Now for the two elements Beam of truss that represent the beam from joint G to H, and calling GH the internal force, we have:

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Now to the element Force that represent the reaction in A:

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And to the two elements Force that represent the reactions in H:

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Now to the elements Force that represent the load forces, three of 600 lb and two of 300 lb:

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Now click in the icon Solve to get the answer:

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