Statics

$H = h + \frac{lMg/2}{\sqrt{4T^2 - (Mg)^2}} \approx 5.5$ m.

The resultant of the cable tension forces must be equal to the weight of the lantern.

No. In this case, the tension force of the rope would be infinite.

$F = kMg/(\cos \alpha + k \sin \alpha) \approx 162$ N.

$\alpha = \arctan k$ (see figure).

$F = \dfrac{kMg}{\cos \alpha + k \sin \alpha}$ (see problem 3).

The expression on the right-hand side is minimized when the value of the denominator $\cos \alpha + k \sin \alpha$ is maximized.

Denoting $k$ through $\tan \beta$, we can transform the expression as follows:

$
\cos \alpha + \tan \beta \sin \alpha = \dfrac{1}{\cos \beta} (\cos \alpha \cos \beta + \sin \alpha \sin \beta) = \dfrac{1}{\cos \beta} \cos (\alpha - \beta).
$

The last expression is maximized when $\cos (\alpha - \beta) = 1$, that is, when $\alpha - \beta = 0$, from which it follows that $\alpha = \beta = \arctan k$.

The action of the moment of force $F_1$ relative to the instantaneous axis will lead to the rotation of the spool around point O clockwise, and it will roll to the right.

The moment of force $F_2$ relative to the instantaneous axis is zero, so in this case, the thread will unwind, leaving the spool in place.

The moment of force $F_3$ will lead to the rotation of the spool counterclockwise, and it will roll to the left.

$F = \dfrac{kmg \tan \alpha}{2(k + \tan \alpha)}$ to move the cart left;

$F' = \dfrac{kmg \tan \alpha}{2(\tan \alpha - k)}$ to move it right if $\tan \alpha > k$. If $\tan \alpha < k$, it's impossible to move the cart right.

The force needed to pull the cart equals the magnitude of the friction force acting on the cart from the rod.

Consider first the case where force $F$ moves the cart left (see figure above). In this case, forces $N$ and $F_{\text{fr}}$ act on the rod, as shown in the figure. Since the rod is in equilibrium, for the moments of forces acting on it, we can write
$\frac{1}{2} mg \sin \alpha - Nl \sin \alpha - F_{\text{fr}} l \cos \alpha = 0$
($l$ is the length of the rod). Since $F_{\text{fr}} = kN$, i.e., $N = F_{\text{fr}}/k$, from this equation we find that
$F = F_{\text{fr}} = \frac{mg \sin \alpha}{2(\cos \alpha + \frac{1}{k} \sin \alpha)} = \frac{kmg \tan \alpha}{2(k + \tan \alpha)}.$

If now force $F'$ must move the cart right, the friction force acting on the rod is directed right, and from the equilibrium condition of the rod, we get
$\frac{1}{2} mg \sin \alpha - Nl \sin \alpha + F'{\text{fr}} l \cos \alpha = 0.$
Assuming the cart moves, i.e.,
$F'{\text{fr}} = kN$ or
$N = F'{\text{fr}}/k,$
we find
$F' = F'{\text{fr}} = \frac{mg \sin \alpha}{2(\sin \alpha - k \cos \alpha)} = \frac{kmg \tan \alpha}{2(\tan \alpha - k)}.$
But if $\tan \alpha < k$, $F'{\text{fr}} \to \infty$. The same happens if $\tan \alpha < k$. From the equilibrium equation of the rod, it follows that $F'{\text{fr}} = (N - mg/2) \tan \alpha$. If $\tan \alpha < k$, the friction force $F'{\text{fr}} < kN - (mg/2)k$, i.e., $F'{\text{fr}} < kN$. This means that if $\tan \alpha < k$, the friction force does not reach the value $kN$ for any value of force $F'$ applied to the cart, and it's impossible to move the cart. 'Jamming' occurs.

$T = mg \dfrac{l+R}{\sqrt{l^2 + 2lR}}$; $F = mg \dfrac{R}{\sqrt{l^2 + 2lR}}$.

For equilibrium of the body, two conditions must be met:

1. the sum of all forces acting on the body must be zero:
   $\sum \vec{F}_i = 0 \tag{1}$
   (this equality must hold for the projections of all forces onto any line);
2. the sum of the moments of all forces acting on the body relative to any fixed axis must also be zero:
   $\sum M_i = 0.$
   As the fixed axis (see figure below), let's choose a horizontal line parallel to the wall and passing through the center of the sphere. Forces $\vec{F}$ and $m\vec{g}$ are directed along the radii of the sphere, so their moments relative to the chosen axis are zero. From the last condition, it follows that the moment of force $\vec{T}$ must also be zero, which means this force must be directed along the radius of the sphere. Let's now write equation (1) for the projections of forces onto the vertical and horizontal directions:
   $mg - T \cos \alpha = 0,$
   $F - T \sin \alpha = 0,$
   where $\sin \alpha = R/(l+R)$, whence we find:
   $T = mg/\cos \alpha = mg(l+R)/\sqrt{l^2+2lR};$
   $F = T \sin \alpha = mg \frac{R}{\sqrt{l^2+2lR}}.$
   According to Newton's third law, the pressure force of the sphere on the wall is equal in magnitude to force F and directed in the opposite direction.

$h = 2r \cot \alpha$.

The cylinder does not tip over if the vertical line passing through the center of gravity of the cylinder passes through its base.

Four dynamometers are sufficient for weighing. The sum of their readings equals the weight of the bar. The metal bar can be suspended from dynamometers located, for example, at the same distance from each other.

$k \ge \tan 15^\circ \approx 0.268$.

The force pushing the wedge out is the resultant of the reaction forces $\vec{F}_1$ and $\vec{F}_2$ of the log (see figure) (due to symmetry $|\vec{F}_1| = |\vec{F}_2|$), and the balancing force is the resultant of the friction forces $F_{fr1} = kF_1$ and $F_{fr2} = kF_2$, directed along the wedge (the force of gravity compared to these forces can be neglected).

$F = Mg/2 \approx 6 \times 10^3$ N.

2754 N on the front wheels and 3861 N on the rear wheels.

a) $\Delta x = mg/k$; $\Delta l = 2\Delta x$;

b) $\Delta x = mg/(2k)$; $\Delta l = \Delta x$.

$k_{eq} = k_1 k_2 / (k_1 + k_2)$ for series connection of springs and $k_{eq} = k_1 + k_2$ for parallel connection.

In series connection of springs, the forces stretching them are equal and equal to the force F stretching the system of springs, and the extension of the system is equal to the sum of the extensions of the springs: $\Delta x = \Delta x_1 + \Delta x_2.$ Since $F = \Delta x k_{eq}$, then $\Delta x = F/k_{eq}.$ Similarly, $\Delta x_1 = F/k_1$ and $\Delta x_2 = F/k_2$. Therefore $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$ and $k_{eq} = \frac{k_1 k_2}{k_1 + k_2}.$ In parallel connection of springs $\Delta x = \Delta x_1 = \Delta x_2$, and $F = F_1 + F_2$. Substituting $F = \Delta x k_{eq}$, $F_1 = \Delta x k_1$ and $F_2 = \Delta x k_2$ here, we get $k_{eq} = k_1 + k_2.$

$l_1/l_2 = 11/13$.

Let $l_1$ and $l_2$ be the lengths of the parts into which the inextensible spring is divided by point A, where a mass $2m$ needs to be suspended, and $k_1$ and $k_2$ be the stiffnesses of these parts of the spring. Since when suspending a weight, the extension of the whole spring is $n$ times greater than the extension of the $1/n$ part of this spring, the stiffness of the whole spring is $n$ times less than the stiffness of the $1/n$ part of the spring (see problem 14), i.e., $k_1 = lk/l_1\quad\text{and}\quad k_2 = lk/l_2,$ where $k = mg/(0.1l)$. When suspending a mass $2m$, a weight $3m$ will act on the upper part of the spring and it will extend by $\Delta l_1 = \frac{3mg}{k_1} = 3mg \frac{l_1 } {lk}.$ The lower part of the spring will extend by $\Delta l_2 = mg/k_2 = mg \frac{l_2}{lk},$ while $l_1 + \Delta l_1 = l_2 + \Delta l_2,$ i.e., $l_1 (\frac{3mg}{lk} + 1) = l_2 (\frac{mg}{lk} + 1).$ Hence $\frac{l_1}{l_2} = \frac{mg+lk}{3mg+lk} = \frac{mg + mg/0.1}{3mg + mg/0.1} = \frac{11}{13}.$

$F > Mg/2$, $k_{min} = 1/2$.

Apply the second condition (see problem 7), choosing the edge about which the cube rotates as the fixed axis. The maximum friction force must be no less than the force F applied to the body.

$F_{min} = Mg\sqrt{2}/4$; $k \ge 1/3$.

The force required to tip the cube is minimal if it is perpendicular to the diagonal plane passing through the edge about which the cube tips, and is applied to the second edge of the cube lying in this diagonal plane.

Solution: If a horizontal force $F=kmg$ is applied to the block near its base, it will start moving. If this force is applied at height $h$ from the floor, then $Fh = mga/2$ and it will start tipping over around the edge of the base. Therefore $kmgh = mga/2$ and $k = a/(2h)$; $a$ is the side of the base.

$\tan \alpha = 1/3$.

The rod will hang such that the vertical line passing through the suspension point A of the rod passes through its center of gravity O (see figure). Let the length of the rod be $2l$, find the lengths of the segments: $AB = l/2 + DB = l/2 + l/4 = \frac{3l}{4}$ and $OB = l/4.$

$x = l \frac{M(n-1)+2mn}{2m(n+1)}$.

The extensions of the springs must be the same. The force $F_1 = \frac{Mg}{2} + \frac{l-x}{l} mg$ acts on the left spring at equilibrium, and the force $F_2 = \frac{Mg}{2} + \frac{x}{l} mg$ acts on the right spring (see figure above).

$F_1 = mg \sin \beta / \sin(\alpha + \beta) = 20$ H; $F_2 = mg \sin \alpha / \sin(\alpha + \beta) = 34$ H.

As fixed axes, it is convenient to choose horizontal lines passing through the points of contact of the sphere with the planes.

$F \ge mg \frac{\sqrt{h(2R-h)}}{R-h}$.

For the wheel to roll over the step, the moment of the force $\vec{F}$ applied to the wheel relative to the axis passing through point A (see figure) must be greater than or equal to the moment of the gravity force $m\vec{g}$ of the wheel relative to the same axis.

To the rim of the wheel. In this case, the moment of the applied force is twice as large.

During braking, the friction force $F_{\text{fr}}$ acts on the car wheels. Let's apply forces $\vec{F}_1$ and $\vec{F}_2$ to the center of gravity of the car, equal to $F_{\text{fr}}$ and directed in different directions (see figure). This can be done, since the sum of these forces and their moments relative to any axis are zero.

Now $\vec{F}1$ can be considered as the braking force, and $F{\text{fr}}$ and $\vec{F}_2$ as a couple of forces causing the car to rotate around the axis passing through the point of contact of the wheels with the road, as a result of which the front end of the car lowers.

The couple of forces will cause the plate to rotate around its center of mass.

Since the sum of the forces acting on the plate is zero, its center of mass will remain at rest.

$F = 0.5 Mg \cos \alpha$; decreases.

$N_A = \dfrac{mg}{2} \cot \alpha$; $Q = \dfrac{mg}{2} \sqrt{4 + \cot^2 \alpha}$.

Let's write the equilibrium conditions for the ladder (see problem 7; see figure).

1st condition: $x$-axis: $N_A - F_{\text{fr}} = 0$; y-axis: $N_B - mg = 0$;

2nd condition: for moments of forces relative to the horizontal axis passing through the points of contact of the ladder with the support,
$N_A l \sin \alpha = mgl \cos \alpha / 2.$

The resultant of forces $N_B$ and $F_{\text{fr}}$ is
$Q = mg/2 \sqrt{4 + \cot^2 \alpha}.$

Force $Q$ must pass through point O of intersection of the lines passing through forces $N_A$ and $mg$, since the sum of the moments of all forces relative to this point must be zero (see figure).

$\sin(\alpha/2) = m_1/m$; $\alpha = 60^\circ$; $F_{AB} = mg \cos^2(\alpha/2) \approx 36.75$ N; unstable equilibrium.

To determine the angle $\alpha$, consider the moments of the forces acting on the rod $Q=m_1 g$ and $mg$ relative to point A (see figure above). Having considered the moments of the forces acting on the rod relative to point C, find the force $F_A$ acting in the hinge A (see problem 27; the line of action of force $F_A$ passes through the intersection point of the lines of action of forces $Q$ and $mg$, lying on the midpoint of side BC of the equilateral triangle ABC): $F_A l \sin(\alpha/2) = mgl \sin \alpha / 2 \implies F_A = mg \cos(\alpha/2).$ Its projection onto the direction of the rod $F_{AB} = F_A \cos(\alpha/2)$. The equilibrium is unstable: for any small deviation from the found position of equilibrium, the system cannot return to it on its own, moreover, the changed ratio of the moments of forces acting on the rod will contribute to the further displacement of the system in the same direction.

$\alpha_{min} = \arctan \frac{(1-k_1 k_2)}{2k_2} \approx 38^\circ 40'$.

See the solution to problem 27. The forces acting on the ladder are shown in the following figure. The limiting angle $\alpha_{min}$, which the ladder can form with the horizon without slipping, will correspond to the maximum possible static friction forces $F_A$ and $F_{\text{fr}}$, i.e., they must simultaneously have the largest possible values, equal respectively to $k_1 N_A$ and $k_2 N_B$.

$h = l F_{\text{fr}} \sin \alpha \tan \alpha / (mg) \approx 2$ m.

See problem 27. The forces acting on the ladder are shown below.

For $k \ge 1/2 (\cot \alpha - 1)$.

Write down the condition for the sum of moments of the reaction forces of the wall and the force of gravity relative to the axis passing through the line of contact of the cube with the floor to be zero, and take into account that the reaction force of the wall must be equal in magnitude to the friction force $kN_1$, and the force $N_1$ is the reaction force of the floor, equal to the force of gravity.

1. Will be disturbed.

2. The mass of the thick end of the rod is greater.

Hint: By lines parallel to any of the sides, the triangle is divided into thin strips. Obviously, the center of gravity of the triangle lies on the line passing through the midpoints of these strips, i.e., on the median.

The center of mass of each rod lies at its midpoint. Connecting the midpoints of the rods, we get a triangle whose side lengths are half the side lengths of the given triangle and, consequently, proportional to the masses of the corresponding rods. The angle bisectors of the resulting triangle divide its sides into parts proportional to the other two sides.

At a distance of 30 cm from the 10 g ball.

The system will be balanced if a force equal to the sum of the weights of the balls is applied to the center of mass of the system. Write down the condition for the sum of moments of all forces acting on the system relative to one of the endpoints to be zero, find the position of the center of mass.

$R/6$ from the center.

Let the distance from the center of mass of the plate to the center of the circle be $x$. From symmetry considerations, it is obvious that the center of mass of the plate lies on its diameter passing through the center of the hole (see figure). If we fill the hole with the material from which the plate is made, the center of the resulting solid plate will be at the center of the circle. Therefore, the equality $m_1 R/2 = x m_2$ must hold. Substituting $m_1 = \pi (R/2)^2 \rho = \pi R^2 \rho / 4$ and $m_2 = \pi R^2 \rho - m_1 = 3/4 \pi R^2 \rho$ here, where $\rho$ is the surface density of the plate material, we get $x = R/6$.

$a\sqrt{3}/28$ from the geometric center of the cube.

See the solution to the previous problem.

Horizontal forces are equal to $mgR/\sqrt{l^2 - 4R^2} \approx 0.16$ N. Vertical force is $mg \approx 0.25$ N.

$M \ge 2m(1 - r/R)$.

The cylinder will not tip over if the moment of its weight $Q$ relative to point $O$ (see figure) is not less than the tipping moment of forces $\vec{F}_1$ and $\vec{F}_2$ of the balls' pressure on the cylinder: $F_1 AO - F_2 B_1 O \le MgR\quad (Mg = Q).\tag{1}$ These forces, according to Newton's third law, are equal in magnitude to the forces $\vec{N}_1$ and $\vec{N}_2$ acting on the balls. On the upper ball, besides force $\vec{F}_1$, act the weight $m\vec{g}$ and the reaction force of the lower ball $\vec{N}_1$, directed along the line of centers. The forces acting on the lower ball are shown in the figure. From the equilibrium condition of the balls, we find that $F_1 = F_2 = mg \tan \alpha.$ Substituting these values into inequality (1) and considering that $AO - B_1 O = 2(R-r) \cot \alpha$, we find that $Mg \ge 2mg(1 - r/R).$

$k \ge 1/(2+\sqrt{3})$.

The logs will be in equilibrium if the moment of the pressure force from the upper log on each of the lower logs does not exceed the moment of the friction force of the upper log with the corresponding lower log relative to the line of contact of the log with the ground.

The cylinder will roll down at an angle of inclination of the plank $\alpha = \arcsin(1/6)$.

The center of mass of the cylinder with the hole (point O) is located at a distance $r = R/6$ from the axis of the cylinder (see problem 36). When the plank is tilted, the cylinder rotates so that its center of mass is on the vertical line passing through the point of contact of the cylinder with the plank. However, if this vertical line passes from the axis of the cylinder at a distance greater than $r=R/6$ (see the following figure), equilibrium is impossible. This means that the cylinder will start rolling when the angle $\alpha$ is such that $\sin \alpha_1 = r/R = (1/6)R/R = 1/6;\quad \alpha_1 = \arcsin(1/6).$ At the same time, the cylinder will start sliding at such an angle of inclination of the plank at which $\tan \alpha_2 = k = 1/5,$ hence $\alpha_2 = \arctan(1/5)$; $\alpha_1 < \alpha_2$, therefore rolling will occur before sliding.

$x = R/\sqrt{2}$.

The system is in indifferent equilibrium if its center of gravity coincides with the geometric center of the hemisphere.

The center of gravity of the glass with water is in the lowest position when it coincides with the water level in the glass. (Explain!)

$x \ge \frac{a}{2} (\frac{\tan \alpha}{k} - 1)$.

The block will not slide out if the component of the gravity force along the block does not exceed the maximum friction force (see figure): $mg \sin \alpha \le F_{fr.max} = N_A k + N_B k$. (1) The pressure forces $N_A$ and $N_B$ of the rods on the block are found from the condition that the sum of the moments of forces relative to points A and B is zero, respectively: $mg x \cos \alpha - N_B a = 0$; $mg(a-x)\cos \alpha - N_A a = 0$. Substituting the values of $N_A$ and $N_B$ found from here into inequality (1), we find that the block will not slide out if $x \ge \frac{a}{2} (\frac{\tan \alpha}{k} - 1)$.

a) 410 J; b) 1100 J.

The minimum work required to be done during infinitely slow tipping of the cube is equal to the difference in potential energies of the cube in the initial position and in the position of unstable equilibrium.

$A_1 = mgka$; $A_2 = 0.207 mga$ for $k=0.207$ (see hint for problem 45).