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# How to draw FBD in physics

Learn how to draw FBD in Physics by following these steps.

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## Step 1: Draw the coordinate axes.

Some authors prefer to draw the z-axis parallel to the FBD, but it is unnecessary.

## Step 2: Sketch the free body diagram.

It must correspond to an equilibrium problem. For example, here is an FBD for a block on a plane inclined at an angle \theta with respect to the horizontal. The tension force in the string is equal to mg minus F fric, where F fric is the frictional force that acts on the block. The tension in the string is equal to the F norm, where the F norm is the normal force N exerted by the inclined plane on the same side of the block as the F fric.

## Step 3: Sketch free-body diagrams for each object shown in step 2, but do not include internal forces.

For example, in the FBD for the block, there is no necessary force shown on any particular side of the block because all forces acting on this block are exerted by other objects.

## Step 4: Solve for unknown quantities using trigonometric or vector identities.

The right-hand rule might be useful in some cases. Note that the angle is usually measured between the direction of movement and the FBD vector and not between two vectors.

## Step 5: Draw free-body diagrams for each object, including forces internal to these objects.

For example, there are two types of internal forces acting on the block: one of these is a normal force at the bottom, and the other is the weight of the block.

## Step 6: Draw FBD for each object again, including all forces acting on them.

In our example, three types of external forces act on the block: gravitational force mg, tension T in the string, and the normal force N exerted by the inclined plane on the same side of the block as F fric. The third FBD has an additional force, labeled “F contact,” which is a force that may be exerted between the block and the table.

## Step 7: Solve for unknown quantities using trigonometric or vector identities.

The right-hand rule might be useful in some cases. Note that the angle is usually measured between the direction of movement and the FBD vector, not between two vectors.

## Step 8: Draw free-body diagrams for each object again, including all forces acting on them.

In our example, five external forces are acting on the block: gravitational force mg, tension T in the string, normal force N exerted by the inclined plane on the same side of the block as F fric, contact force F contact, which is a force that may be exerted between the block and surface it rests upon, and friction force F fric. There are also internal forces within this object: one is a normal force at the bottom, and another is the weight of the block.

## Step 9: Solve for unknown quantities using trigonometric or vector identities.

The right-hand rule might be useful in some cases. Note that the angle is usually measured between the direction of movement and the FBD vector rather than between two vectors.

## Step 10: Draw free-body diagrams for each object again, including all forces acting on them.

In our example, five external forces are acting on the block: gravitational force mg, tension T in a string which is equal to F norm, normal force N exerted by the inclined plane on the same side of the block as F fric, contact force F contact which is a force that may be exerted between the block and surface it rests upon, and friction force F fric. There are also internal forces within this object: one is a normal force at the bottom, and another is the weight of the block.

## Step 11: How to draw FBD continued

Check that steps 8 and 10 both give the same answer for unknown quantities, and that step 9 gives the correct angle between two vectors in question. In our example, we find that F norm = mg – F fric = mg – (mg cos \theta – N), so T in string = mg cos \theta – (mg cos \theta – N). Solving for θ yields tan(θ) = (F norm /F contact )/(mgcos\theta-N), which reduces to tan(\theta) = (mgcos\theta-N)/F norm . This is the same equation as in step 8. F contact = mg sin \theta.

## Step 12: When all calculations are complete, write your results.

If your answer was not one of the three listed possibilities for θ (0°, 90°, or 180°), then show all necessary steps to justify how you determined that angle–you may need to redo this whole procedure if you do so! In our problem, T in string = 81 N and θ = 71°. Notice that because the block rests on a surface other than the inclined plane itself, it experiences two types of external force: gravitational force mg acting down the plane’s slope and contact force F contact acting out of the page. We find that the block’s acceleration is g sin (71°) = 0.69 m/s2, which means it would require 69 N to hold this block at rest on an incline like this one.

## Step 13: Produce a free-body diagram showing all forces acting on your object after solving the problem.

Next step in how to draw FBD is here. Explain any changes in direction or magnitude of forces if they exist compared to before solving. Our FBD after solving contains only five external forces: gravitational force mg, tension T in the string, normal force N exerted by the inclined plane on the same side of the block as F fric, contact force F contact, which is a force that may be exerted between the block and surface it rests upon, and friction force F fric. Note that the direction of gravity is different than before solving. Our object’s weight vector was pointing more downward into the hillside while standing on the slope.

## Step 14: Expand your free-body diagram to show your object’s displacement, if any.

In this case, it would be up the plane, which isn’t shown in this sketch but might be more obvious in another one. After solving, we do not include any forces related to movement in our FBD, so our object does not experience a displacement.

## Step 15: Draw a second free-body diagram showing all forces acting on your object after solving the problem.

Explain any changes in direction or magnitude of forces if they exist compared to before solving. Here we show our FBD after solving five external forces. First is gravitational force mg down the plane’s slope. Contact force F contact, which is a force that may be exerted between the block and surface it rests upon, and friction force F fric. Note that gravity’s direction is different from before solving because while standing on the slope, our object’s weight vector was pointing more downward into the hillside than when lying flat at rest on the inclined plane. It means that when standing on an incline like this one, a person’s weight is not vertical down the plane’s slope.

## Step 16: Solve for unknown quantities in problems

stated with equilibrium conditions by putting the given information into your chosen equation to solve, paying attention to signs (+ or -) and direction of each force (tension, contact force, etc.). The equation that we use here is F norm = ma. We can rearrange this equation to find acceleration a = F norm /m. Now plug all known values into these equations and solve for unknowns! T in string = 81 N , m= 65 kg , mg cos \theta = 853 N , N= 390 N . Using these values, we can solve for θ using the equation T in string = mg sin \theta. Substituting these known values into this equation, we get (65 kg)a=853 N or a=-0.80 m/s2. The negative sign means that our block accelerates to the right (downhill).

## Step 17:

Solve for unknown quantities in problems stated with equilibrium conditions by putting the given information into your chosen equation to solve, paying attention to signs (+ or -) and direction of each force (tension, contact force, etc.). The equation that we use here is F norm = ma. We can rearrange this equation to find acceleration = F norm /m. Now plug all known values into these equations and solve for unknowns! T in string = 81 N , m= 65 kg , F fric = 38 N , mg cos \theta = 853 N . Using these values, we can solve for θ using the equation T in string – mg sin \theta = F fric . Substituting these values into this equation, we get (65 kg)-(853 N )sin\theta=81N-mgcos\theta, or sin\theta=-0.32 . If we assume that sin\theta=-1, then our block accelerates down the plane at a=3.38 m/s2. Note that angles we express angles less than 90° using a negative sine value.

How to draw FBD here? We could also solve the object’s displacement by summing up its initial displacement and its final velocity. Δx=v I +v f. We would get v i =0 and v f =0.38 m/s or Δx=1.38 meters. It is in the direction of gravity (down the plane). We are assuming that the object does not experience any other forces during this time period such as friction, contact force, etc.