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Two blocks on an inclined plane

Construct the free-body diagram for object A and object B in [link] .

Strategy

We follow the four steps listed in the problem-solving strategy.

Solution

We start by creating a diagram for the first object of interest. In [link] (a), object A is isolated (circled) and represented by a dot.

Figure a shows two objects on an inclined plane, sloping down to the left. Object A is on top of object B. A free body diagram shows T pointing right and up, parallel to the plane, N subscript BA pointing left and up, perpendicular to the plane, f subscript BA pointing left and down, parallel to the plane and w subscript A pointing vertically down. W subscript A is weight of block A, T is tension, N subscript BA is normal force exerted by B on A, f subscript BA is friction force exerted by B on A. Figure b shows the objects on the slope in the same manner. A free body diagram has f subscript B and f subscript AB pointing right and up, parallel to the slope, N subscript B pointing left and up perpendicular to the slope, w subscript B pointing vertically down and N subscript AB pointing down and right, perpendicular to the slope. W subscript B is weight of block B, N subscript AB is normal force exerted by A on B, N subscript B is normal force exerted by the incline plane on B. f subscript AB is friction force exerted by A on B. f subscript B is friction force exerted by the incline plane on B.
(a) The free-body diagram for isolated object A. (b) The free-body diagram for isolated object B. Comparing the two drawings, we see that friction acts in the opposite direction in the two figures. Because object A experiences a force that tends to pull it to the right, friction must act to the left. Because object B experiences a component of its weight that pulls it to the left, down the incline, the friction force must oppose it and act up the ramp. Friction always acts opposite the intended direction of motion.

We now include any force that acts on the body. Here, no applied force is present. The weight of the object acts as a force pointing vertically downward, and the presence of the cord indicates a force of tension pointing away from the object. Object A has one interface and hence experiences a normal force, directed away from the interface. The source of this force is object B, and this normal force is labeled accordingly. Since object B has a tendency to slide down, object A has a tendency to slide up with respect to the interface, so the friction f BA is directed downward parallel to the inclined plane.

As noted in step 4 of the problem-solving strategy, we then construct the free-body diagram in [link] (b) using the same approach. Object B experiences two normal forces and two friction forces due to the presence of two contact surfaces. The interface with the inclined plane exerts external forces of N B and f B , and the interface with object B exerts the normal force N AB and friction f AB ; N AB is directed away from object B, and f AB is opposing the tendency of the relative motion of object B with respect to object A.

Significance

The object under consideration in each part of this problem was circled in gray. When you are first learning how to draw free-body diagrams, you will find it helpful to circle the object before deciding what forces are acting on that particular object. This focuses your attention, preventing you from considering forces that are not acting on the body.

Two blocks in contact

A force is applied to two blocks in contact, as shown.

Strategy

Draw a free-body diagram for each block. Be sure to consider Newton’s third law at the interface where the two blocks touch.

Two squares are shown in contact with each other. The left one is smaller and is labeled m1. The right one is bigger and is labeled m2. An arrow pointing right towards m1 is labeled F.

Solution

Figure shows two free body diagrams. The first one shows arrow A subscript 21 pointing left and arrow F pointing right. The second one shows arrow A 12 pointing right. Both diagrams also have arrows pointing up and down.

Significance

A 21 is the action force of block 2 on block 1. A 12 is the reaction force of block 1 on block 2. We use these free-body diagrams in Applications of Newton’s Laws .

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Block on the table (coupled blocks)

A block rests on the table, as shown. A light rope is attached to it and runs over a pulley. The other end of the rope is attached to a second block. The two blocks are said to be coupled. Block m 2 exerts a force due to its weight, which causes the system (two blocks and a string) to accelerate.

Strategy

We assume that the string has no mass so that we do not have to consider it as a separate object. Draw a free-body diagram for each block.

Figure shows block m1 placed on a table. A string attached to it runs over a pulley and down the right side of the table. A block m2 is suspended from it. An arrow a1 points right and an arrow a2 points down.

Solution

Figure a shows block m1. An arrow labeled m1g point upwards from it, an arrow N points downwards and an arrow T points right. Figure b shows block m2. An arrow T points upwards from it and an arrow m2g points downwards.

Significance

Each block accelerates (notice the labels shown for a 1 and a 2 ); however, assuming the string remains taut, they accelerate at the same rate. Thus, we have a 1 = a 2 . If we were to continue solving the problem, we could simply call the acceleration a . Also, we use two free-body diagrams because we are usually finding tension T , which may require us to use a system of two equations in this type of problem. The tension is the same on both m 1 and m 2 .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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