All instances in this chapter are planar problems. Accordingly, we usage equilibrium problems in the component form of (Figure) to (Figure). We introduced a problem-solving strategy in (Figure) to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a perform of procedures to follow once solving revolution equilibrium problems for expanded rigid bodies. We proceed in 5 practical steps.

You are watching: Find t, the tension in the cable. remember to account for all the forces in the problem.

Problem-Solving Strategy: static EquilibriumIdentify the thing to be analyzed. For some equipment in equilibrium, it may be essential to consider an ext than one object. Identify all pressures acting ~ above the object. Recognize the inquiries you should answer. Recognize the information provided in the problem. In realistic problems, some an essential information might be implicitly in the instance rather than noted explicitly.Set up a free-body diagram for the object. (a) pick the

*xy*-reference framework for the problem. Draw a free-body diagram because that the object, including only the forces that plot on it. Once suitable, stand for the forces in regards to their components in the preferred reference frame. Together you perform this because that each force, cross out the original force so the you execute not erroneously include the same pressure twice in equations. Label all forces—you will need this because that correct computations of net pressures in the

*x*– and also

*y*-directions. Because that an unknown force, the direction need to be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The exactly direction is identified by the authorize that you obtain in the last solution. A add to sign

way that the working direction is the really direction. A minus authorize

way that the actual direction is opposite come the assumed functioning direction. (b) choose the ar of the rotation axis; in other words, pick the pivot suggest with respect to which you will compute torques of exhilaration forces. Top top the free-body diagram, show the place of the pivot and also the bar arms of exhilaration forces—you will require this because that correct computations of torques. In the an option of the pivot, keep in mind that the pivot deserve to be placed everywhere you wish, however the guiding rule is the the best selection will simplify as lot as feasible the calculation of the net torque follow me the rotation axis.Simplify and solve the device of equations because that equilibrium to attain unknown quantities. At this point, her work requires algebra only. Save in mind the the variety of equations should be the exact same as the variety of unknowns. If the variety of unknowns is bigger than the variety of equations, the trouble cannot be solved.

Note that setup up a free-body diagram because that a rigid-body equilibrium difficulty is the most important component in the solution process. Without the correct setup and also a exactly diagram, you will certainly not be able to write under correct problems for equilibrium. Additionally note the a free-body diagram for an extended rigid human body that might undergo rotational activity is different from a free-body diagram for a body that experiences only translational motion (as you observed in the chapters top top Newton’s legislations of motion). In translational dynamics, a body is stood for as that is CM, wherein all pressures on the body are attached and also no torques appear. This go not organize true in rotational dynamics, where an extended rigid body cannot be represented by one allude alone. The factor for this is that in evaluating rotation, we must identify torques exhilaration on the body, and torque depends both ~ above the acting force and on its lever arm. Here, the free-body diagram for an extensive rigid body helps united state identify external torques.

### Example

The torque BalanceThree masses space attached to a uniform meter stick, as displayed in (Figure). The mass of the meter stick is 150.0 g and also the masses come the left that the fulcrum room

and also

uncover the mass

that balances the device when the is attached in ~ the right finish of the stick, and the common reaction pressure at the fulcrum as soon as the mechanism is balanced.

**Figure 12.9**In a talk balance, a horizontal beam is sustained at a fulcrum (indicated by S) and also masses space attached to both political parties of the fulcrum. The mechanism is in revolution equilibrium when the beam does not rotate. That is well balanced when the beam stays level.

Strategy

For the arrangement shown in the figure, we determine the following 5 forces exhilaration on the meter stick:

is the load of fixed

is the weight of mass

is the load of the entire meter stick;

is the weight of unknown massive

is the regular reaction force at the support suggest *S*.

We choose a framework of referral where the direction the the *y*-axis is the direction of gravity, the direction of the *x*-axis is follow me the meter stick, and also the axis that rotation (the *z*-axis) is perpendicular come the *x*-axis and passes with the support point *S*. In various other words, we select the pivot in ~ the point where the meter stick touches the support. This is a natural selection for the pivot due to the fact that this suggest does not move as the pole rotates. Currently we are ready to collection up the free-body diagram because that the meter stick. We show the pivot and attach five vectors representing the five forces follow me the heat representing the meter stick, locating the pressures with respect to the pivot (Figure). At this stage, we can identify the bar arms of the 5 forces given the information noted in the problem. For the three hanging masses, the difficulty is explicit about their locations along the stick, however the information about the ar of the weight *w* is provided implicitly. The an essential word below is “uniform.” We know from ours previous researches that the cm of a uniform pole is located at the midpoint, therefore this is wherein we affix the weight *w*, at the 50-cm mark.

**Figure 12.10**Free-body diagram for the meter stick. The pivot is preferred at the support allude S.

Solution

With (Figure) and also (Figure) because that reference, we start by finding the bar arms of the five forces acting on the stick:

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this method the 2nd equilibrium problem is

We settle these equations simultaneously for the unknown values

and also

In (Figure), we cancel the *g* factor and rearrange the state to obtain

To find the common reaction force, we rearrange the terms in (Figure), converting grams to kilograms:

Significance

Notice the (Figure) is independent of the worth of *g*. The torque balance may thus be used to measure up mass, due to the fact that variations in *g*-values on earth’s surface perform not affect these measurements. This is no the situation for a feather balance since it procedures the force.

Repeat (Figure) making use of the left end of the meter stick to calculate the torques; that is, by place the pivot at the left end of the meter stick.

316.7 g; 5.8 N

In the following example, we show how to use the first equilibrium condition (equation because that forces) in the vector form given by (Figure) and (Figure). We present this equipment to illustrate the importance of a suitable an option of referral frame. Although every inertial reference frames space equivalent and also numerical solutions acquired in one structure are the exact same as in any kind of other, an unsuitable selection of reference structure can make the systems quite an extensive and convoluted, whereas a wise an option of reference frame makes the systems straightforward. We show this in the identical solution come the exact same problem. This details example illustrates an applications of static equilibrium come biomechanics.

### Example

Forces in the ForearmA weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) v his forearm, as presented in (Figure). His forearm is positioned at

v respect to his top arm. The forearm is sustained by a contraction of the biceps muscle, which reasons a torque about the elbow. Assuming the the stress in the biceps acts along the vertical direction given by gravity, what tension should the muscle exert to organize the forearm at the position shown? What is the pressure on the elbow joint? Assume the the forearm’s weight is negligible. Offer your last answers in SI units.

**Figure 12.11**The forearm is rotated approximately the elbow (E) through a contraction of the biceps muscle, which reasons tension

Strategy

We determine three pressures acting ~ above the forearm: the unknown pressure

in ~ the elbow; the unknown stress and anxiety

in the muscle; and also the load

with magnitude

We adopt the structure of referral with the *x*-axis follow me the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, i beg your pardon is the exact same as the direction the the upper arm. The *x*-axis makes an angle

through the vertical. The *y*-axis is perpendicular come the *x*-axis. Now we set up the free-body diagram because that the forearm. First, we attract the axes, the pivot, and the 3 vectors representing the three established forces. Then we find the edge

and also represent each pressure by that *x*– and *y*-components, remembering come cross the end the original force vector to avoid twin counting. Finally, we label the forces and also their lever arms. The free-body diagram because that the forearm is displayed in (Figure). In ~ this point, us are all set to collection up equilibrium problems for the forearm. Each force has *x*– and *y*-components; therefore, we have two equations because that the very first equilibrium condition, one equation for each component of the net force acting on the forearm.

Notice that in our frame of reference, contributions to the 2nd equilibrium problem (for torques) come just from the *y*-components of the forces because the *x*-components that the forces are every parallel to their lever arms, so that for any of castle we have

in (Figure). For the *y*-components we have

in (Figure). Also notice that the torque of the pressure at the elbow is zero due to the fact that this pressure is attached at the pivot. Therefore the contribution to the network torque comes only from the torques the

and of

Solution

We watch from the free-body diagram that the *x*-component that the net pressure satisfies the equation

(Figure) and (Figure) are two equations that the first equilibrium problem (for forces). Next, we read from the free-body diagram the the network torque along the axis of rotation is

(Figure) is the 2nd equilibrium problem (for torques) for the forearm. The free-body diagram shows that the lever arms space

and also

in ~ this point, we do not require to convert inches right into SI units, because as lengthy as these devices are continual in (Figure), lock cancel out. Utilizing the free-body diagram again, we discover the magnitudes the the component forces:

When we simplify these equations, we see that we space left with just two live independence equations for the 2 unknown pressure magnitudes, *F* and *T*, because (Figure) for the *x*-component is tantamount to (Figure) for the *y*-component. In this way, we attain the first equilibrium condition for forces

and the 2nd equilibrium condition for torques

The magnitude of stress and anxiety in the muscle is acquired by resolving (Figure):

The force at the elbow is acquired by resolving (Figure):

The negative sign in the equation tells united state that the actual force at the elbow is antiparallel come the functioning direction adopted for illustration the free-body diagram. In the last answer, we transform the forces into SI systems of force. The price is

Significance

Two important problems here room worth noting. The an initial concerns conversion into SI units, which have the right to be done at the an extremely end that the equipment as lengthy as we store consistency in units. The 2nd important issue comes to the hinge joints such together the elbow. In the initial evaluation of a problem, hinge joints should always be assumed come exert a pressure in an *arbitrary direction*, and then you need to solve because that all contents of a hinge force independently. In this example, the elbow force happens to it is in vertical since the difficulty assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a basic rule.

Suppose we embrace a reference frame with the direction the the *y*-axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only *y*-components, for this reason we have actually only one equation for the first equilibrium problem (for forces). We attract the free-body diagram because that the forearm as shown in (Figure), indicating the pivot, the acting forces and also their bar arms v respect come the pivot, and the angles

and

the the forces

and

(respectively) make v their lever arms. In the meaning of torque offered by (Figure), the angle

is the direction edge of the vector

count *counterclockwise* native the radial direction the the lever arm that always points far from the pivot. Through the very same convention, the angle

is measured *counterclockwise* from the radial direction of the lever arm to the vector

excellent this way, the non-zero torques space most quickly computed by directly substituting right into (Figure) as follows:

**Figure 12.13**Free-body diagram because that the forearm for the identical solution. The pivot is situated at point E (elbow).

The second equilibrium condition,

have the right to be now written as

From the free-body diagram, the first equilibrium problem (for forces) is

(Figure) is the same to (Figure) and also gives the an outcome

(Figure) gives

We check out that this answers are identical to our previous answers, yet the 2nd choice because that the structure of referral leads to an equivalent solution that is simpler and quicker since it go not need that the pressures be resolved into their rectangular components.

### Example

A Ladder Resting versus a WallA uniform ladder is

long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as displayed in (Figure). The inclination angle between the ladder and the stormy floor is

find the reaction pressures from the floor and also from the wall on the ladder and also the coefficient of revolution friction

in ~ the interface of the ladder through the floor that avoids the ladder native slipping.

Strategy

We deserve to identify four forces acting top top the ladder. The first force is the typical reaction force *N* from the floor in the upward vertical direction. The 2nd force is the revolution friction pressure

directed horizontally along the floor toward the wall—this force stays clear of the ladder from slipping. This two pressures act top top the ladder in ~ its contact allude with the floor. The third force is the load *w* of the ladder, attached at its CM situated midway in between its ends. The fourth force is the normal reaction force *F* from the wall surface in the horizontal direction away from the wall, attached in ~ the contact suggest with the wall. There space no other forces due to the fact that the wall surface is slippery, which means there is no friction in between the wall surface and the ladder. Based upon this analysis, we embrace the framework of referral with the *y*-axis in the upright direction (parallel come the wall) and the *x*-axis in the horizontal direction (parallel to the floor). In this frame, each pressure has either a horizontal ingredient or a vertical component yet not both, i beg your pardon simplifies the solution. We pick the pivot at the contact allude with the floor. In the free-body diagram because that the ladder, we show the pivot, all 4 forces and also their lever arms, and the angles in between lever arms and also the forces, as displayed in (Figure). Through our an option of the pivot location, there is no torque one of two people from the regular reaction pressure *N* or indigenous the static friction *f* because they both act in ~ the pivot.

where

is the talk of the weight *w* and

is the torque of the reaction *F*. Indigenous the free-body diagram, we identify that the bar arm that the reaction in ~ the wall surface is

and also the lever arm that the weight is

v the aid of the free-body diagram, we identify the angle to be supplied in (Figure) for torques:

because that the torque from the reaction force with the wall, and

because that the torque because of the weight. Now we are prepared to use (Figure) to compute torques:

We acquire the common reaction force with the floor by addressing (Figure):

The magnitude of friction is acquired by fixing (Figure):

The coefficient of static friction is

The net pressure on the ladder at the contact point with the floor is the vector amount of the normal reaction from the floor and also the static friction forces:

We need to emphasize here two basic observations of handy use. First, notification that once we pick a pivot point, over there is no expectation the the device will actually pivot around the favored point. The ladder in this instance is not rotating in ~ all yet firmly stand on the floor; nonetheless, that is contact point with the floor is a an excellent choice for the pivot. Second, an alert when we use (Figure) for the computation of separation, personal, instance torques, we execute not have to resolve the forces into their normal and also parallel components with respect come the direction that the bar arm, and also we execute not require to consider a sense of the torque. As long as the edge in (Figure) is properly identified—with the aid of a free-body diagram—as the edge measured counterclockwise native the direction the the bar arm to the direction that the pressure vector, (Figure) provides both the magnitude and also the sense of the torque. This is since torque is the vector product the the lever-arm vector crossed v the force vector, and also (Figure) expresses the rectangle-shaped component of this vector product along the axis of rotation.

SignificanceThis result is independent of the length of the ladder because *L* is cancelled in the second equilibrium condition, (Figure). No matter exactly how long or short the ladder is, as lengthy as its load is 400 N and also the angle with the floor is

our results hold. However the ladder will slip if the net torque becomes an unfavorable in (Figure). This wake up for some angles when the coefficient of revolution friction is not great enough to prevent the ladder from slipping.

For the instance described in (Figure), identify the worths of the coefficient

of revolution friction because that which the ladder starts slipping, offered that

is the angle the the ladder makes with the floor.

A swinging door the weighs

is supported by hinges *A* and *B* so the the door deserve to swing around a upright axis passing with the hinges (Figure). The door has actually a width of

and the door slab has actually a uniform mass density. The hinges are inserted symmetrically in ~ the door’s leaf in together a means that the door’s load is same distributed between them. The hinges are separated by street

find the pressures on the hinges once the door rests half-open.

Solution

From the free-body diagram because that the door we have the very first equilibrium condition for forces:

We choose the pivot at point *P* (upper hinge, per the free-body diagram) and also write the second equilibrium condition for torques in rotation about point *P*:

In assessing

we use the geometry that the triangle displayed in part (a) of the figure. Now we substitute this torques right into (Figure) and also compute

The forces on the hinges are discovered from Newton’s 3rd law as

Solve the difficulty in (Figure) by acquisition the pivot position at the facility of mass.

### Check your Understanding

Is it possible to remainder a ladder versus a rough wall when the floor is frictionless?

A painter climbs a ladder. Is the ladder more likely to slip once the painter is near the bottom or near the top?

### Problems

The uniform seesaw shown listed below is well balanced on a fulcrum located 3.0 m from the left end. The smaller boy ~ above the right has a fixed of 40 kg and also the bigger young on the left has a fixed 80 kg. What is the mass of the board?

A uniform 40.0-kg frame of length 6.0 m is sustained by 2 light cables, as shown below. One 80.0-kg painter stands 1.0 m indigenous the left end of the scaffold, and also his painting equipment is 1.5 m native the ideal end. If the tension in the left cable is double that in the appropriate cable, find the tensions in the cables and also the fixed of the equipment.

To obtain up top top the roof, a person (mass 70.0 kg) areas a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the basic of the ladder 2.00 m from the house. The ladder rests versus a plastic rain gutter, i beg your pardon we have the right to assume to it is in frictionless. The center of massive of the ladder is 2.00 m indigenous the bottom. The person is stand 3.00 m from the bottom. Discover the typical reaction and friction pressures on the ladder at its base.

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A uniform horizontal strut weighs 400.0 N. One finish of the strut is attached to a hinged assistance at the wall, and also the other end of the strut is attached come a sign that weighs 200.0 N. The strut is likewise supported by a cable attached in between the end of the strut and the wall. Assuming that the whole weight of the sign is attached in ~ the really end the the strut, discover the tension in the cable and also the force at the hinge that the strut.

The forearm shown listed below is positioned at an angle

with respect to the top arm, and a 5.0-kg mass is held in the hand. The total mass of the forearm and hand is 3.0 kg, and also their facility of fixed is 15.0 cm from the elbow. (a) What is the magnitude of the force that the biceps muscle exerts on the forearm because that

(b) What is the magnitude of the pressure on the elbow joint because that the exact same angle? (c) how do these forces depend on the angle

The uniform boom shown below weighs 700 N, and also the thing hanging native its right end weighs 400 N. The boom is supported by a light cable and also by a hinge at the wall. Calculate the stress and anxiety in the cable and also the force on the hinge ~ above the boom. Walk the pressure on the hinge act follow me the boom?

over the horizontal; no

A uniform trapdoor shown below is 1.0 m by 1.5 m and also weighs 300 N. It is sustained by a solitary hinge (H), and by a light rope tied in between the middle of the door and also the floor. The door is hosted at the position shown, where its slab makes a