第12章静的つり合いと弾性

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure 12.11. 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.

We identify three forces acting on the forearm: the unknown force $\overrightarrow{F}$ at the elbow; the unknown tension ${\overrightarrow{T}}_{\text{M}}$ in the muscle; and the weight $\overrightarrow{w}$ 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 12.12. 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.

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

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.

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 12.13, 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 ${\overrightarrow{T}}_{\text{M}}$ and $\overrightarrow{w}$ (respectively) make with their lever arms. In the definition of torque given by Equation 12.10, the angle ${\theta }_T$ is the direction angle of the vector ${\overrightarrow{T}}_{\text{M}},$ 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 $\overrightarrow{w}.$ Done this way, the non-zero torques are most easily computed by directly substituting into Equation 12.10 as follows:

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 12.14. The minimum inclination angle between the ladder and the rough floor below which the ladder slips 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.

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 12.15. 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.

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

This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, Equation 12.31. 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 Equation 12.31. This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

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 12.16. 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.

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 $\overrightarrow{A}$ from hinge $A,$ an unknown force $\overrightarrow{B}$ from hinge $B,$ and the known weight $\overrightarrow{w}$ 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 12.17, and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force $\overrightarrow{A}$ $(A_x$ and $A_y),$ and two components of force $\overrightarrow{B}$ $(B_x$ 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 12.17. Finally, we solve the equations for the unknown force components and find the forces.

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

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.