Example

The Torque Balance

Three masses are attached to a uniform meter stick, as shown in (Figure). The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are $$ {m}_{1}=50.0\,\text{g} $$ and $$ {m}_{2}=75.0\,\text{g}. $$ Find the mass $$ {m}_{3} $$ that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure 12.9 In a torque balance, a horizontal beam is supported at a fulcrum (indicated by S) and masses are attached to both sides of the fulcrum. The system is in static equilibrium when the beam does not rotate. It is balanced when the beam remains level.

Strategy

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

$${w}_{1}={m}_{1}g $$ is the weight of mass $$ {m}_{1}; $$ $${w}_{2}={m}_{2}g $$ is the weight of mass $$ {m}_{2};$$

$$w=mg $$ is the weight of the entire meter stick; $$ {w}_{3}={m}_{3}g $$ is the weight of unknown mass $$ {m}_{3};$$

$${F}_{S} $$ is the normal reaction force at the support point S.

We choose a frame of reference where the direction of the y-axis is the direction of gravity, the direction of the x-axis is along the meter stick, and the axis of rotation (the z-axis) is perpendicular to the x-axis and passes through the support point S. In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot (Figure). At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w, at the 50-cm mark.

Figure 12.10 Free-body diagram for the meter stick. The pivot is chosen at the support point S.

Solution

With (Figure) and (Figure) for reference, we begin by finding the lever arms of the five forces acting on the stick:

Now we can find the five torques with respect to the chosen pivot:

The second equilibrium condition (equation for the torques) for the meter stick is

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

Selecting the $$ +y$$-direction to be parallel to $$ {\overset{\to }{F}}_{S}, $$ the first equilibrium condition for the stick is

Substituting the forces, the first equilibrium condition becomes

We solve these equations simultaneously for the unknown values $$ {m}_{3} $$ and $$ {F}_{S}. $$ In (Figure), we cancel the g factor and rearrange the terms to obtain

To obtain $$ {m}_{3} $$ we divide both sides by $$ {r}_{3}, $$ so we have

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

Significance

Notice that (Figure) is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

Example

Forces in the Forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in (Figure). His forearm is positioned at $$ \beta =60\text{°} $$ with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

Figure 12.11 The forearm is rotated around the elbow (E) by a contraction of the biceps muscle, which causes tension $$ {\overset{\to }{T}}_{\text{M}}.$$

Strategy

We identify three forces acting on the forearm: the unknown force $$ \overset{\to }{F} $$ at the elbow; the unknown tension $$ {\overset{\to }{T}}_{\text{M}} $$ in the muscle; and the weight $$ \overset{\to }{w} $$ with magnitude $$ w=50\,\text{lb}. $$ We adopt the frame of reference with the x-axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x-axis makes an angle $$ \beta =60\text{°} $$ with the vertical. The y-axis is perpendicular to the x-axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle $$ \beta $$ and represent each force by its x– and y-components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in (Figure). At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x– and y-components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Figure 12.12 Free-body diagram for the forearm: The pivot is located at point E (elbow).

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y-components of the forces because the x-components of the forces are all parallel to their lever arms, so that for any of them we have $$ \text{sin}\,\theta =0 $$ in (Figure). For the y-components we have $$ \theta =±90\text{°} $$ in (Figure). Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of $$ {T}_{y} $$ and of $$ {w}_{y}.$$

Solution

We see from the free-body diagram that the x-component of the net force satisfies the equation

and the y-component of the net force satisfies

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

(Figure) is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are $$ {r}_{T}=1.5\,\text{in}\text{.} $$ and $$ {r}_{w}=13.0\,\text{in}\text{.} $$ At this point, we do not need to convert inches into SI units, because as long as these units are consistent in (Figure), they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into (Figure), (Figure), and (Figure) to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T, because (Figure) for the x-component is equivalent to (Figure) for the y-component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques

The magnitude of tension in the muscle is obtained by solving (Figure):

The force at the elbow is obtained by solving (Figure):

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

Significance

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction, and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Solution

Suppose we adopt a reference frame with the direction of 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, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in (Figure), indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles $$ {\theta }_{T} $$ and $$ {\theta }_{w} $$ that the forces $$ {\overset{\to }{T}}_{\text{M}} $$ and $$ \overset{\to }{w} $$ (respectively) make with their lever arms. In the definition of torque given by (Figure), the angle $$ {\theta }_{T} $$ is the direction angle of the vector $$ {\overset{\to }{T}}_{\text{M}}, $$ counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle $$ {\theta }_{w} $$ is measured counterclockwise from the radial direction of the lever arm to the vector $$ \overset{\to }{w}. $$ Done this way, the non-zero torques are most easily computed by directly substituting into (Figure) as follows:

Figure 12.13 Free-body diagram for the forearm for the equivalent solution. The pivot is located at point E (elbow).

The second equilibrium condition, $$ {\tau }_{T}+{\tau }_{w}=0, $$ can be now written as

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

(Figure) is identical to (Figure) and gives the result $$ T=433.3\,\text{lb}. $$ (Figure) gives

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Example

A Ladder Resting Against a Wall

A uniform ladder is $$ L=5.0\,\text{m} $$ long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in (Figure). The inclination angle between the ladder and the rough floor is $$ \beta =53\text{°}. $$ Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction $$ {\mu }_{\text{s}} $$ at the interface of the ladder with the floor that prevents the ladder from slipping.

Figure 12.14 A 5.0-m-long ladder rests against a frictionless wall.

Strategy

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force $$ f={\mu }_{\text{s}}N $$ directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y-axis in the vertical direction (parallel to the wall) and the x-axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in (Figure). With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

Figure 12.15 Free-body diagram for a ladder resting against a frictionless wall.

Solution

From the free-body diagram, the net force in the x-direction is

the net force in the y-direction is

and the net torque along the rotation axis at the pivot point is

where $$ {\tau }_{w} $$ is the torque of the weight w and $$ {\tau }_{F} $$ is the torque of the reaction F. From the free-body diagram, we identify that the lever arm of the reaction at the wall is $$ {r}_{F}=L=5.0\,\text{m} $$ and the lever arm of the weight is $$ {r}_{w}=L\,\text{/}\,2=2.5\,\text{m}. $$ With the help of the free-body diagram, we identify the angles to be used in (Figure) for torques: $$ {\theta }_{F}=180\text{°}-\beta $$ for the torque from the reaction force with the wall, and $$ {\theta }_{w}=180\text{°}+(90\text{°}-\beta ) $$ for the torque due to the weight. Now we are ready to use (Figure) to compute torques:

We substitute the torques into (Figure) and solve for $$ F:$$

We obtain the normal reaction force with the floor by solving (Figure): $$ N=w=400.0\,\text{N}. $$ The magnitude of friction is obtained by solving (Figure): $$ f=F=150.7\,\text{N}. $$ The coefficient of static friction is $$ {\mu }_{\text{s}}=f\,\text{/}\,N=150.7\,\text{/}\,400.0=0.377.$$

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

Its magnitude is

and its direction is

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use (Figure) for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in (Figure) is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, (Figure) gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and (Figure) expresses the rectangular component of this vector product along the axis of rotation.

Significance

This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, (Figure). No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is $$ 53\text{°}, $$ our results hold. But the ladder will slip if the net torque becomes negative in (Figure). This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

Example

Forces on Door Hinges

A swinging door that weighs $$ w=400.0\,\text{N} $$ is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges (Figure). The door has a width of $$ b=1.00\,\text{m}, $$ and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance $$ a=2.00\,\text{m}. $$ Find the forces on the hinges when the door rests half-open.

Figure 12.16 A 400-N swinging vertical door is supported by two hinges attached at points A and B.

Strategy

The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force $$ \overset{\to }{A} $$ from hinge $$ A, $$ an unknown force $$ \overset{\to }{B} $$ from hinge $$ B, $$ and the known weight $$ \overset{\to }{w} $$ attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y-axis along the direction of gravity and the x-axis in the plane of the slab, as shown in panel (a) of (Figure), and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force $$ \overset{\to }{A} $$ $$({A}_{x} $$ and $$ {A}_{y}), $$ and two components of force $$ \overset{\to }{B} $$ $$({B}_{x} $$ and $$ {B}_{y}). $$ In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns $$ ({A}_{x}, $$ $${B}_{x}, $$ $${A}_{y}, $$ and $$ {B}_{y}), $$ we must set up four independent equations. One equation is the equilibrium condition for forces in the x-direction. The second equation is the equilibrium condition for forces in the y-direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, $$ {A}_{y}={B}_{y}. $$ To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of (Figure). Finally, we solve the equations for the unknown force components and find the forces.

Figure 12.17 (a) Geometry and (b) free-body diagram for the door.

Solution

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

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

We use the free-body diagram to find all the terms in this equation:

In evaluating $$ \text{sin}\,\beta , $$ we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into (Figure) and compute $$ {B}_{x}:$$

Therefore the magnitudes of the horizontal component forces are $$ {A}_{x}={B}_{x}=100.0\,\text{N}. $$ The forces on the door are

The forces on the hinges are found from Newton’s third law as

Significance

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.