Corrected exercises: static equilibrium (203-SN1-RE)

First condition: $\sum\vec{F}=\vec{0}$, which rules out any acceleration of the CENTRE OF MASS, and therefore any translation. Second condition: $\sum\tau=0$, which rules out any angular acceleration, and therefore any rotation. Both are necessary and independent.

The sum of the forces is indeed ZERO, since the two forces are equal and opposite. Yet the bar is NOT in equilibrium: applied at different points, the two forces turn the bar in the SAME direction. Taking the centre of the bar as the pivot, each force has a lever arm of 0.400 m and produces a torque of $20.0\times 0.400=8.00$ N·m, both in the same sense: $\tau_{net}=16.0$ N·m. This is called a COUPLE, and it is exactly what you apply to a steering wheel or a screwdriver.

Take two pivots $O$ and $O'$, and let $\vec{d}$ be the vector from $O$ to $O'$. The torque of a force $\vec{F}_{i}$ applied at $P_{i}$ equals, about $O'$, the torque about $O$ reduced by the torque that $\vec{F}_{i}$ would produce if applied at $O$ with the lever arm $\vec{d}$. Summing over all the forces, the extra term factors as $\vec{d}$ multiplied by $\sum\vec{F}_{i}$. If this sum is zero, the term vanishes and the two sums of torques are EQUAL.

You may choose the pivot FREELY, and you should take advantage of that. The smart choice is the point of application of an unknown force, because its lever arm becomes zero and it drops out of the torque equation. A problem with two unknown reactions is then solved with a single equation in one unknown instead of a system. This is the central technique of the whole chapter, and it explains why the result of part c is not a theoretical curiosity but the main working tool.

Take the left support as the pivot: its reaction $R_{1}$ has a zero lever arm and drops out. The torques of the weights are clockwise, that of $R_{2}$ is counterclockwise. $R_{2}\times 6.00=200\times 9.80\times 3.00+500\times 9.80\times 2.00=5880+9800=15{,}680$ N·m, so $R_{2}=\dfrac{15{,}680}{6.00}=2613$ N.

First method, using the sum of the forces: $R_{1}+R_{2}=\left(200+500\right)\times 9.80=6860$ N, so $R_{1}=6860-2613=4247$ N. Second method, using torques about the RIGHT support: $R_{1}\times 6.00=200\times 9.80\times 3.00+500\times 9.80\times 4.00=5880+19{,}600=25{,}480$ N·m, so $R_{1}=4247$ N. The two agree, which validates the whole calculation. Note that the support CLOSEST to the load carries the larger share.

By symmetry, the two reactions are equal if the load is at the MIDPOINT, at 3.00 m. Each then equals $\dfrac{6860}{2}=3430$ N. We can check this: since the beam is uniform, its own weight is already shared equally; the load only has to be centred as well.

The left support stops being loaded when $R_{1}=0$, that is, when the sum of the torques about the right support vanishes without it. This would require the torque of the load to offset that of the beam's weight in the opposite direction, which is impossible as long as the load stays BETWEEN the supports: the two torques then have the same sense. The load would have to be placed beyond the right support, as an overhang. The condition becomes $500\times 9.80\times x=200\times 9.80\times 3.00$ with $x$ measured beyond the right support, giving $x=1.20$ m. Beyond that, $R_{1}$ would become negative: since a simple support can only PUSH, the beam TIPS. This is the calculation that sizes the counterweights of a crane.

A distributed load exerts an elementary weight on each slice. The sum of these weights equals the total weight, and the sum of their torques about any point is the same as that of a single force applied at the centre of gravity of the distribution. Replacing the distributed load by this resultant is therefore exact, not approximate, for any equilibrium calculation. For a uniform beam, this point is the midpoint: $W=60.0(9.80)=588$ N applied at $x=3.00$ m. If the load were triangular, the resultant would sit at the centre of gravity of the triangle, that is TWO THIRDS of the length from the end where the load is zero, and not at the midpoint.

Let $R_{A}$ and $R_{B}$ be the reactions at $x=0$ and $x=4.00$ m. Torques about $A$: $R_{B}(4.00)=588(3.00)$, so $R_{B}=441$ N. Sum of the vertical forces: $R_{A}=588-441=147$ N. The overhang already tilts the load toward the right support: it carries three times as much as the left one, even though the beam is uniform.

Take torques about $B$ and require $R_{A}=0$. The weight of the beam, at $x=3.00$ m, is on the left side of $B$ and therefore tends to bring the left part back down: its torque is $588(1.00)=588$ N$\cdot$m. The worker, at $x$, must produce the opposite torque: $784\left(x-4.00\right)=588$, so $x-4.00=0.750$ and $x=4.75$ m. Beyond $4.75$ m, the beam tips about $B$ if it is merely resting on its supports: the left support lifts off. If it is bolted down, it must instead PULL downward, and it is the fastening that does the work.

Torques about $A$: $R_{B}(4.00)=588(3.00)+784(6.00)=1{,}764+4{,}704=6{,}468$, so $R_{B}=1{,}617$ N. Sum of the forces: $R_{A}=588+784-1{,}617=-245$ N. The negative sign does not mean a calculation error: it indicates that the force assumed to point upward actually points DOWNWARD. The left support must hold the beam down with $245$ N, which a simple resting support cannot do. A simple support can only PUSH, like the bottom of a bucket or the ground under a crate.

The counterweight, placed at $x=0$, must produce about $B$ a torque that offsets that of the worker. Torques about $B$: $m g(4.00)+588(1.00)=784(2.00)$, so $4.00\,mg=1{,}568-588=980$ and $mg=245$ N, hence $m=25.0$ kg. This is exactly the $245$ N calculated in part d), which is consistent. In practice, much more would be used, for three reasons: this value corresponds to STRICT equilibrium, where the slightest disturbance tips the beam; the worker may carry a load, jump or lean, which increases the torque; and every building code imposes a factor of safety. An equilibrium calculated to the newton is an equilibrium you do not build.

Rectangle 1, the base: area $A_{1}=2.00\times 0.50=1.00$ m², centre at $\left(1.00,\ 0.25\right)$. Rectangle 2, the upright: area $A_{2}=0.50\times 1.80=0.90$ m², centre at $\left(0.25,\ 0.50+0.90\right)=\left(0.25,\ 1.40\right)$, since it rests on the first.

$x_{G}=\dfrac{A_{1}x_{1}+A_{2}x_{2}}{A_{1}+A_{2}}=\dfrac{1.00\times 1.00+0.90\times 0.25}{1.90}=\dfrac{1.225}{1.90}=0.645$ m. $y_{G}=\dfrac{1.00\times 0.25+0.90\times 1.40}{1.90}=\dfrac{1.51}{1.90}=0.795$ m. The centre of gravity is therefore at $\left(0.645,\ 0.795\right)$.

Because the plate is UNIFORM and of constant thickness: the mass of each piece is then proportional to its area, $m_{i}=\sigma A_{i}$, with the same mass per unit area $\sigma$ for all of them. In the quotient that defines $x_{G}$, the factor $\sigma$ appears in both the numerator and the denominator, and cancels. The argument would fail if the two parts were made of different materials.

Yes, and it is common. The point $\left(0.645,\ 0.795\right)$ found in part b falls precisely in the notch of the L, where there is no material. Other examples: the centre of a ring, of a horseshoe, of a boomerang. The centre of gravity is a point of calculation, not a piece of the object: nothing requires material to be there.

Resting on its base, the plate is stable: the vertical through $x_{G}=0.645$ m falls well inside the base, which extends from 0 to 2.00 m. Tilting the plate about its right edge, located at $x=2.00$ m, it tips when the vertical through the centre of gravity passes beyond that edge. The critical angle satisfies $\tan\theta=\dfrac{2.00-0.645}{0.795}=1.704$, that is $\theta=59.6^{\circ}$. Tipping the other way, to the left about the edge $x=0$, the critical angle is much smaller: $\tan\theta=\dfrac{0.645}{0.795}=0.811$, that is $39.1^{\circ}$. The plate is therefore clearly less stable on the side of the upright, which is intuitive since the mass is shifted there.

Four forces: the weight of the ladder, $18.0\times 9.80=176.4$ N applied at the midpoint, 2.50 m from the foot; the weight of the person, $75.0\times 9.80=735$ N at 3.50 m; the reaction of the wall; the reaction of the floor, which has a normal component and a friction component. Since the wall is SMOOTH, it cannot exert any tangential force along its surface, hence no vertical component: its reaction is necessarily HORIZONTAL, perpendicular to the wall.

Pivot at the foot: the two floor reactions have a zero lever arm there and drop out. The lever arm of a weight, a vertical force, is the HORIZONTAL distance to the pivot, that is $\ell\cos 65^{\circ}$ for a point at distance $\ell$ along the ladder. The lever arm of the wall force, horizontal, is the height $5.00\sin 65^{\circ}$. Hence $N_{wall}\times 5.00\sin 65^{\circ}=176.4\times 2.50\cos 65^{\circ}+735\times 3.50\cos 65^{\circ}$. The calculation gives $N_{wall}\times 4.532=186.4+1087.3=1273.7$, so $N_{wall}=281$ N.

Vertically, the floor supports the entire weight, since the wall pushes only horizontally: $N_{floor}=176.4+735=911$ N. Horizontally, the friction of the floor exactly balances the push of the wall: $f=N_{wall}=281$ N. The no-slip condition is written $f\leq\mu_{s}N_{floor}$, so $\mu_{s}\geq\dfrac{281}{911}=0.308$.

The general condition, obtained by redoing the calculation of part b with an arbitrary angle $\theta$, is $\tan\theta\geq\dfrac{m_{L}\frac{L}{2}+m_{P}d}{\mu_{s}\left(m_{L}+m_{P}\right)L}$. Numerically, $\tan\theta\geq\dfrac{18.0\times 2.50+75.0\times 3.50}{0.400\times 93.0\times 5.00}=\dfrac{307.5}{186.0}=1.653$, so $\theta\geq 58.8^{\circ}$. The ladder inclined at 65.0° is therefore well above the limit: it holds.

Because the destabilizing torque grows with the distance $d$ from the person to the foot, whereas the resisting torque does not change. Climbing therefore increases $N_{wall}$, and so the friction required at the floor, without increasing the normal reaction beyond the weight already counted. A ladder that is not steep enough makes everything worse, because the lever arm of the weights varies as $\cos\theta$, which grows as $\theta$ decreases, while the lever arm of the wall reaction varies as $\sin\theta$, which shrinks. Both effects work in the same direction and the ratio blows up. That is why the safety rule is to place the foot at a quarter of the ladder's length from the wall, which corresponds to about 75°, and never to climb onto the top rungs.

Weight: $W=80.0(9.80)=784$ N. Sliding: it begins when the force reaches the maximum static friction, $F_{slide}=\mu_{s}W=0.500(784)=392$ N. Tipping: the wardrobe pivots about its lower front edge, and the threshold is reached when the torque of the push equals that of the weight, whose lever arm is the half-width: $F\,h=W\frac{w}{2}$, that is $F_{tip}=\frac{784(0.300)}{1.20}=196$ N. It is the SMALLER of the two thresholds that decides: the wardrobe tips at $196$ N, long before it could slide. The mover who pushes too high overturns the furniture instead of moving it along.

The two thresholds coincide when $\mu_{s}W=\frac{Ww}{2h_{c}}$, which gives $h_{c}=\frac{w}{2\mu_{s}}$. The weight cancels: the mass of the wardrobe has no influence on what it does, only on the force required. Here $h_{c}=\frac{0.600}{2(0.500)}=0.600$ m. Below this height it slides, above it it tips. Pushing low makes it slide; pushing high overturns it.

With $\mu_{s}=0.200$, the critical height becomes $h_{c}=\frac{0.600}{2(0.200)}=1.50$ m. The push is applied at $1.20$ m, that is BELOW $h_{c}$: this time the wardrobe slides, from $F=0.200(784)=157$ N, whereas $196$ N would still be needed to tip it. The same gesture, at the same place, produces two opposite behaviours depending on the floor covering. Waxing a floor protects furniture from tipping over, but makes it slide away more easily.

Take torques about the centre $O$ of the base. The weight acts directly above $O$: its torque is zero. Friction acts at floor level, at zero height: its torque about $O$ is zero as well. That leaves the push, with torque $F h=150(1.20)=180$ N$\cdot$m, and the normal force, whose line of action has shifted by a distance $d$ toward the front edge: $W d=180$, so $d=\frac{180}{784}\approx 0.230$ m. Since the half-width is $0.300$ m, the normal force stays within the base and equilibrium holds, which confirms the $196$ N threshold. What the calculation reveals is that the normal force is not a fixed force applied at the centre: it is the resultant of a pressure distributed over the whole base, and this resultant MIGRATES toward the edge as the push increases. Tipping occurs exactly when it reaches the edge: there is no material beyond to carry it.

On a curve, the force that plays the role of the push is the centripetal force $\frac{mv^{2}}{R}$, and it is applied at the centre of mass, at height $h_{cm}$. Tipping: $\frac{mv^{2}}{R}h_{cm}\geq mg\frac{w}{2}$, that is $\frac{v^{2}}{Rg}\geq\frac{w}{2h_{cm}}$. Sliding: $\frac{v^{2}}{Rg}\geq\mu_{s}$. The vehicle rolls over before skidding if $\frac{w}{2h_{cm}}<\mu_{s}$. Truck: $\frac{2.40}{2(1.80)}=0.667$, less than $0.800$, so it TIPS. Sports car: $\frac{1.80}{2(0.450)}=2.00$, far greater than $0.800$, so it SKIDS, and by a wide margin. The whole art of a racing chassis lies in this ratio: a wide track and a centre of mass close to the ground. And it explains why a very grippy road surface, far from protecting the truck, increases its risk of rollover by letting it reach the tipping threshold.

At equilibrium, the torques are equal: $F_{in}\times 0.600=200\times 0.150$, so $F_{in}=\dfrac{30.0}{0.600}=50.0$ N. The mechanical advantage is 4: the force produced is four times the force applied, at the price of a displacement four times larger.

Pivot at the elbow, which eliminates the unknown joint force. The torque of the biceps, counterclockwise, balances those of the two weights, clockwise: $F_{b}\times 0.050=5.00\times 9.80\times 0.350+1.50\times 9.80\times 0.150=17.15+2.205=19.36$ N·m. Hence $F_{b}=\dfrac{19.355}{0.050}=387$ N.

The weight lifted is only $5.00\times 9.80=49.0$ N. The biceps must therefore pull about EIGHT TIMES harder than the load carried. This is a third-class lever, where the driving force is applied BETWEEN the pivot and the load: there is no mechanical advantage, only a disadvantage. The muscle must make up for a lever arm seven times shorter than that of the load.

The vertical forces must balance. Upward, the biceps pulls 387.1 N. Downward, the two weights total $49.0+14.7=63.7$ N. The difference, $387.1-63.7=323.4$ N, must be balanced by the joint, which therefore exerts on the forearm a force of 323 N directed DOWNWARD. The elbow is thus compressed far more strongly than the load held would suggest, which explains joint wear in manual workers.

Because a lever always trades force and displacement, and the opposite compromise would be unworkable. With an arm of 0.050 m against 0.350 m, the hand moves SEVEN TIMES faster and seven times farther than the attachment point of the biceps. A muscle can shorten by only about a third of its length: if the attachment were farther out to gain force, the range of the movement would collapse and the arm could barely bend. Nature therefore chose SPEED and RANGE OF MOTION, even if that demands very powerful muscles. It is the right compromise for throwing, striking or catching, activities where speed matters more than raw force.

Theorem: a rigid body in equilibrium under the action of only THREE non-parallel forces necessarily has its three lines of action concurrent at a single point. Proof: let $P$ be the intersection point of the lines of action of two of them, say $\vec{F}_{1}$ and $\vec{F}_{2}$; their torques about $P$ are zero, since their lever arm there is zero. The second condition of equilibrium requires the sum of the torques about $P$ to be zero, so the torque of $\vec{F}_{3}$ about $P$ is zero too. A nonzero force with zero torque about $P$ necessarily has its line of action passing through $P$. The three lines therefore meet at $P$.

Place $A$ at the origin. End $B$ is at $\left(2.00\cos 55.0^{\circ},\ 2.00\sin 55.0^{\circ}\right)=\left(1.147,\ 1.638\right)$ m and the centre of gravity at the midpoint, at $\left(0.574,\ 0.819\right)$ m. The weight is vertical: its line of action is the line $x=0.574$. The rope is horizontal and attached at $B$: its line of action is the line $y=1.638$. They intersect at $P\left(0.574,\ 1.638\right)$. The floor reaction must therefore pass through $A(0,\ 0)$ AND through $P$: its direction makes with the horizontal the angle $\arctan\frac{1.638}{0.574}\approx 70.7^{\circ}$. Note that $\tan 70.7^{\circ}=2\tan 55.0^{\circ}$, because the centre of gravity is at mid-length: the meeting point is twice as high as it is wide relative to the bar. All this without writing an equilibrium equation.

Torques about $A$: the tension, horizontal and applied at $B$, has as its lever arm the height of $B$, that is $1.638$ m; the weight $W=196$ N has as its lever arm the horizontal position of the centre of gravity, that is $0.574$ m. Equilibrium gives $T(1.638)=196(0.574)$, so $T\approx 68.6$ N. Horizontal forces: the floor reaction must balance the tension, so $R_{x}=68.6$ N. Vertical forces: since the rope is horizontal, the floor alone carries the weight, so $R_{y}=196$ N. Magnitude: $R=\sqrt{68.62^{2}+196^{2}}\approx 208$ N, and direction $\arctan\frac{196}{68.62}\approx 70.7^{\circ}$ above the horizontal. This is indeed the direction predicted in part b).

The floor reaction splits into a normal force $N=R_{y}=196$ N and a friction force $f=R_{x}=68.6$ N. We need $f\leq\mu_{s}N$, so $\mu_{s}\geq\frac{68.62}{196}\approx 0.350$. Replacing the values by their expressions, $\mu_{min}=\frac{T}{W}=\frac{W\frac{L}{2}\cos\theta}{W L\sin\theta}=\frac{1}{2\tan\theta}$, and $\frac{1}{2\tan 55.0^{\circ}}\approx 0.350$. Neither the mass nor the length comes in: only the angle decides. The more the bar lies down, the smaller $\tan\theta$ becomes and the grippier the floor must be, which is the same result as for the ladder against a wall.

The lever arm of the tension is halved, since it is now applied at mid-height, at $y=0.819$ m, while the torque of the weight is unchanged. The tension therefore DOUBLES: $T(0.819)=196(0.574)$ gives $T\approx 137$ N. The reaction becomes $R_{x}=137.2$ N and $R_{y}=196$ N, a direction $\arctan\frac{196}{137.2}=55.0^{\circ}$: exactly the angle of the bar. The floor reaction is therefore directed ALONG THE BAR. This is no accident: the meeting point is now $\left(0.574,\ 0.819\right)$, that is, the centre of gravity itself, which belongs to the bar; the line $AP$ is therefore the bar. The required coefficient of friction becomes $\frac{1}{\tan 55.0^{\circ}}\approx 0.700$, twice the previous one. Attaching the rope closer to the foot of the bar is always costly, in tension as well as in grip.

Pivot at the hinge, which eliminates its reaction. Only the PERPENDICULAR component of the cable produces a torque, namely $T\sin 40.0^{\circ}$, with a lever arm of 3.00 m. The two weights are vertical, hence perpendicular to the horizontal boom, with lever arms of 2.00 m and 4.00 m. Hence $T\sin 40.0^{\circ}\times 3.00=60.0\times 9.80\times 2.00+200\times 9.80\times 4.00=1176+7840=9016$ N·m. Then $T=\dfrac{9016}{3.00\times 0.6428}=4675$ N.

The weight of the load is $200\times 9.80=1960$ N: the tension is 2.4 times greater. Two causes add up. First, the cable is attached at 3.00 m while the load hangs at 4.00 m, which gives it a shorter lever arm. Then, and above all, it pulls at an angle of only 40.0°, so the useful fraction of its tension is only $\sin 40^{\circ}=0.643$. A cable nearly parallel to the boom would be almost useless.

Horizontally, only the horizontal component of the cable pulls the boom toward the mast: $H_{x}=T\cos 40.0^{\circ}=4675\times 0.766=3582$ N, the hinge having to push by the same amount in the opposite direction. Vertically, $H_{y}+T\sin 40.0^{\circ}=\left(60.0+200\right)\times 9.80$, so $H_{y}=2548-3005=-457$ N. The total force of the hinge has magnitude $\sqrt{3582^{2}+457^{2}}=3611$ N.

The vertical component is NEGATIVE: the hinge pulls the boom DOWNWARD, it does not support it. This was predictable: the vertical component of the cable, $3005$ N, exceeds on its own the total weight of $2548$ N. The cable therefore lifts more than necessary, and the hinge must hold the boom down to prevent it from pivoting upward. It is a counterintuitive but perfectly common result in this type of setup, and it requires the hinge to be a pin capable of pulling, not a simple support.

It would INCREASE, because the lever arm of the cable would decrease while that of the loads would stay the same: for the same torque, a force applied closer to the pivot must be greater. Attaching the cable at 1.50 m instead of 3.00 m would double the tension, to more than 9300 N. The practical constraint is therefore to attach the cable AS FAR as possible from the hinge and at the widest possible angle, which explains the silhouette of cranes, with their tall masts and widely spread stays. The whole design aims to maximize $\sin\theta$ and the lever arm simultaneously.

The radius is $0.600$ cm $=6.00\times 10^{-3}$ m, so $A=\pi r^{2}=\pi\left(6.00\times 10^{-3}\right)^{2}\approx 1.131\times 10^{-4}$ m$^{2}$. The force is $F=1{,}200(9.80)=1.176\times 10^{4}$ N, so the stress is $\sigma=\frac{F}{A}=\frac{11{,}760}{1.131\times 10^{-4}}\approx 1.04\times 10^{8}$ Pa, that is $104$ MPa. This is about $42\,\%$ of the elastic limit: the cable works within its elastic range and will return exactly to its length once unloaded.

Hooke's law for a solid is written $\sigma=E\varepsilon$, where $\varepsilon=\frac{\Delta L}{L}$ is the strain, a dimensionless quantity. So $\Delta L=\frac{\sigma L}{E}=\frac{\left(1.0398\times 10^{8}\right)(12.0)}{2.00\times 10^{11}}\approx 6.24\times 10^{-3}$ m, that is $6.24$ mm. The strain is $\varepsilon=5.2\times 10^{-4}$, that is, five ten-thousandths: metals deform very little before giving way, which makes them both reassuring and treacherous.

The cabin accelerates upward, so the tension exceeds the weight: $F=m(g+a)=1{,}200(9.80+1.50)=1.356\times 10^{4}$ N. Stress: $\sigma=\frac{13{,}560}{1.131\times 10^{-4}}\approx 1.20\times 10^{8}$ Pa. Elongation: $\Delta L=\frac{\left(1.199\times 10^{8}\right)(12.0)}{2.00\times 10^{11}}\approx 7.19$ mm, nearly a millimetre more. Factor of safety with respect to breaking: $\frac{4.00\times 10^{8}}{1.199\times 10^{8}}\approx 3.34$. A simple start dropped the margin from $3.85$ to $3.34$, a reminder that a cable is never sized on the static load alone.

Elastic limit: $F=\sigma_{el}A=2.50\times 10^{8}\left(1.131\times 10^{-4}\right)\approx 2.83\times 10^{4}$ N, so $m=\frac{28{,}274}{9.80}\approx 2.89\times 10^{3}$ kg. Breaking: $F=4.00\times 10^{8}\left(1.131\times 10^{-4}\right)\approx 4.52\times 10^{4}$ N, so $m\approx 4.62\times 10^{3}$ kg. With a factor of safety of $10$, the allowable load falls to $\frac{4{,}616}{10}\approx 462$ kg. Yet this cable is asked to carry $1{,}200$ kg: on its own it is far outside the code. The conclusion is that of real industry: an elevator is always suspended from SEVERAL cables, each able to hold the cabin by itself, plus a mechanical safety brake. This is not decorative redundancy, it is the only way to respect a factor of $10$ without oversized cables.

The stress is $\frac{F}{A}$ and the area is proportional to the SQUARE of the diameter: doubling the diameter quadruples $A$, hence divides the stress by four, to $26.0$ MPa. The elongation is $\frac{FL}{AE}$: the length doubles, the area quadruples, so $\Delta L$ is divided by two, to $3.12$ mm. Stress and elongation therefore do not follow the same law, which rules out any blanket reasoning. Finally, the elongation depends on the length and the cross-section, that is, on the GEOMETRY of the piece, not only on its material. The intrinsic quantity is Young's modulus $E=\frac{\sigma}{\varepsilon}$, which compares a stress with a relative strain and equals $2.00\times 10^{11}$ Pa for both cables, since they are made of the same steel. Comparing raw elongations means confusing the shape of the piece with the nature of the metal.