Now we will explain the FBD concept, using the following free body diagram example problem as shown in Fig.1. stationary box must be pulled up a 30 degree inclined by a pulley system. The coefficient of static friction between the box and that incline is 0.25. Assuming there is no friction in the pulley system, what force in Newton’s (N) must be applied to the rope in order to move the box up the incline?
Fig. 1: Example of Free Body Diagram
How to draw free body diagram
Step 1: Draw the object with no extra features.
Step 2: Identify the forces acting on the box. The box has mass, so it should also have weight, and a force acting downward. Because the stationary box is on a surface, there is a normal force that acts perpendicular to the surface. Further, attached to the box, there is a rope with tension applied. This tension force will act in the direction of the rope. Since the rope is directly attached to the box and in order to move it up the incline, there will be frictional force impeding movement. This force will then act in the opposite direction, down the incline.
3: Add the forces to the image of the object and label the directions of forces in degrees from the vertical or horizontal axis as understood by the geometry in the example. Refer Fig. 2 that shows above 3 steps.
Step 4: Label all the known values. At this point, weight is known, that is mass (50 kg) multiplied by gravitational constant (9.8 m/s2). The FBD now contains all the given, important information.
Step 5: As a general rule, the Free-Body Diagram has to be oriented, so that the direction of movement is along with one of the principal axes. In this example, the entire diagram could be reoriented by rotating it to 30° counter-clockwise. This step results in the direction of movement that occurs along the x-axis, and this results in three of the four forces that also get oriented along the x or y-axis. Fig. 3 shows Step 4 & Step 5.
Solving the Free-Body Diagram
To solve the problem, the force on the rope required to move the box up the incline should be found. This is called the tension force. Finding this force requires a system of equations. As there is currently one known variable, the weight, there are three unknown variables also; therefore, three equations are required. These equations establish a relationship between all of the forces and are required in order to solve for each force.
In this system of equations, the first one creates a relationship between normal force and frictional force. Since the box will not move until the tension force overcomes the frictional force, the maximum frictional force is needed. The maximum frictional force is equal to the normal force multiplied by the surface coefficient of static friction, that is 0.25, so the first equation will be: Fnorm * 0.25 = Ffrict
As the question refers to the minimum force required to begin moving the object, the system will be in static equilibrium before the time of movement. When a system that is in static equilibrium is concerned, there are two equations that are formed: ∑Fx=0 and ∑Fy=0 or, the sum of all the forces in the y and x-direction should be equal to zero.
When all forces in each direction are summed up, it is noted that any force that acts along either the x or y-axis (in this case tension, frictional and normal force) will only be present in that axis equation. However, weight does not act along one axis, and therefore, should be broken down into its component parts. This we get by using trigonometric functions. In this case, since the weight acts at a 60-degree angle, the vertical component will be Sin 60 degree * W, and the horizontal component will be Cos 60 degree * W.
To achieve the final two equations, the components of the various forces will be added together that acts on each axis. Because of the direction of the positive x and y-axis chosen in the previous step, forces on the x-axis which is acting to the left will be negative, whereas forces that are acting to the right will be positive. Likewise, forces on the y axis which is acting downwards will be negative, and forces that are acting upwards will be positive. Looking at the FBD, three forces will act in the x-direction: the tension force, the horizontal component of the weight, and the frictional force, and two forces will act in the y-direction: the vertical component of the weight and the normal force. This leads to the final equations as follows:
∑Fx= Ffrict+cos(60°)*W–T = 0 or Ffrict + cos(60°)*W=T
∑Fy= Fnorm– sin(60°)*W = 0 or Fnorm= sin(60°)*W
Now that the system of equations has been found, the unknown variables can be found by using known information, that the weight is 490 Newton. Then Equations are combined in order to solve for the tension. Fnorm=sin (60°)*490 N = 424.4 N
424.4 N*0.25 =Ffrict = 106.1 N
106.1 N+cos(60°)*490=T = 351.1 N
With tension force found, the answer to the problem is 351.1 N, that is, this amount of force should be applied to the rope in order to move the box up the incline.
In a similar way, most of the physics or engineering problems can be simplified, visualized, and solved by drawing a free body diagram of the problem. If there is more than one object then a free body diagram of each object can be generated.
