Questions & explanations
1. Two forces P and Q act at an angle of 150°. A third force R keeps them in equilibrium. Express P, Q, R using Lami's theorem.
Let the three forces be P, Q, R with directions such that they balance. The angle between P and Q is 150°, so the angle opposite R is 150°. In the force triangle, the other two angles sum to 30°. But we cannot determine them uniquely without more information? Actually, if only P and Q are given, R is determined by the triangle law: R = √(P²+Q²+2PQ cos150°). But Lami's theorem gives ratios: P/sin(angle opposite P) = Q/sin(angle opposite Q) = R/sin150°. We need the directions of P and Q relative to R. Typically, if we assume P and Q are on either side of R symmetrically or something, we could proceed. A common setup: P and Q are equal and symmetric? Without that, we need the other angles. So, a better phrasing: If the angles between R and P, and R and Q are known (say α and β, with α+β=210°? Wait, sum of all three angles around point is 360°. Actually, the three forces meet at a point, so the angles between them are the angles of the triangle's supplements. The angles of the triangle are not the same as the angles between forces. For Lami, the angle opposite a force is the angle betwee
2. A body is in equilibrium under three forces: 10 N, 20 N, and an unknown force. The angle between the 10 N and 20 N forces is 120°. Find the unknown force using Lami's theorem.
Let the three forces be A=10 N, B=20 N, and C unknown. Label the angles: angle opposite A is between B and C; but we know the angle between A and B is 120°. In a triangle of forces, the interior angle opposite C is the angle between A and B when placed tail to tail? Actually careful: In Lami's theorem, the angle opposite a force is the angle between the other two forces. So, if the angle between 10 N and 20 N is 120°, that angle is opposite the unknown force C. So γ = 120°. Now we need other angles. For equilibrium, the force triangle closes, so sum of angles = 180°. So the other two angles sum to 60°. But we don't have them. This is underdetermined? Wait, Lami's theorem requires all three angles. Actually, if we only know one angle, we cannot find the unknown force uniquely; we need at least one more angle or the triangle shape. Alternatively, we can use law of cosines if we consider vector addition. Maybe the problem is: given magnitude of two forces and the angle between them, we can find the third force by the triangle law: resultant of 10 and 20 at 120° is sqrt(10^2+20^2+2*10*20
3. A real cantilever beam has a point load at its free end. How do you set up the conjugate beam to find the slope and deflection at the free end?
Real beam: fixed left, free right, moment diagram triangular (zero at free, max at fixed). M/EI diagram is a triangle with apex at fixed end. Conjugate beam: free left, fixed right, loaded with this triangular M/EI (upward if real beam sags). To find slope at real free end, compute shear at conjugate right fixed end. To find deflection, compute moment at conjugate right fixed end. The reactions at the fixed end are found from equilibrium of the triangle load. Shear = area of load triangle = PL²/(2EI). Moment = area × distance from right = PL²/(2EI) × L/3 = PL³/(6EI). Wait, standard result is PL³/(3EI), so check: For cantilever with point load at free end, M(x)=–Px (from free end), M/EI = –Px/EI. Area of triangle (0 to L) = –PL²/(2EI). The centroid from fixed end? The triangle is from fixed (max) to free (zero). For conjugate fixed at right, we need moment about that fixed end. If we take right as fixed, triangle base at left? Actually careful: Real beam fixed at left x=0, free at x=L. M(x)=–P(L–x) maybe? Let's not confuse. Standard deflection is PL³/(3EI). The conjugate method yields
4. A force of 10 N acts at point (2,3) meters from the origin. Its components are 6 N along x and 8 N along y. Find its moment about the origin using Varignon's theorem.
Varignon's theorem says the total moment is the sum of moments of components. The moment of a force about a point is force times perpendicular distance. For the 6 N horizontal component, it acts horizontally through point (2,3). Its perpendicular distance from origin vertically is 3 m, so its moment = 6 N × 3 m = 18 Nm. But careful: sign convention: moment tending to rotate counterclockwise is positive. The 6 N force at (2,3) pointing positive x: its line is horizontal at y=3. The perpendicular distance from origin to this line is 3 (vertical). The force tends to rotate clockwise about origin? Let's determine: force to the right, so it would push the point clockwise? Actually, imagine a point at (2,3), force to the right. The moment about origin: position vector (2,3) cross force (6,0) = 2*0 - 3*6 = -18 Nm (clockwise). For 8 N vertical component, line vertical through x=2, distance 2 m, moment = 8 N × 2 m = 16 Nm, but cross product: (2,3) x (0,8) = 2*8 - 3*0 = 16 Nm (counterclockwise). Sum = -18 + 16 = -2 Nm. So moment is 2 Nm clockwise. Simplify: Answer: The 6 N component gives a cl
5. Use the conjugate beam method to describe how to find the maximum deflection of a simply supported beam with a central point load P.
Real beam: length L, point load P at L/2. Bendling moment diagram is triangular, maximum PL/4 at center. M/EI diagram is a triangle of height PL/(4EI). Conjugate beam: simply supported same length, loaded with this symmetric triangle. The reactions at conjugate supports are each half the total area: (1/2)(L/2)(PL/(4EI)) = PL²/(16EI). The slope at the real supports is zero, so conjugate shear at supports is slopes? Actually, for simple support, real slope is not zero. But we want maximum deflection at center. Deflection equals conjugate moment at center. Compute moment in conjugate beam at midspan: reaction times distance minus area of half triangle to center. The half triangle from left support to center has area = (1/2)(L/2)(PL/(4EI)) = PL²/(16EI). Its centroid from left end is 1/3 of the half-span from left, or L/6. But easier: due to symmetry, conjugate moment at center is reaction × L/2 – area of left half-load × distance from center to its centroid. The left half-load is a triangle, area = PL²/(16EI), centroid from left end = L/3 from left, so from center = L/2 – L/3 = L/6. So m
6. Give a simple example of using D'Alembert's principle for a moving object.
Imagine a block being pulled on a smooth floor by a rope with a force of 20 N. The block has mass 5 kg and accelerates at 3 m/s². Using D'Alembert, we add an inertia force of 5 times 3, which is 15 N, opposite to motion. Now we write equilibrium: 20 N pull minus 15 N inertia force equals zero net force? That doesn't match. Wait: correct setup: the net real force is 20 N (if no friction), causing acceleration. Inertia force is -15 N. Then 20 + (-15) = 5 N, not zero. That is wrong. Let's reconsider: D'Alembert's principle states: sum of real external forces + inertia force = 0. So if the real external force is 20 N forward, and we want equilibrium, we need inertia force of -20 N, meaning mass times acceleration is 20 N, so acceleration is 4 m/s². So if the block actually accelerates at 4 m/s², then inertia force is 20 N backwards, and 20 + (-20) = 0. So D'Alembert is used to relate forces to acceleration. A proper example: a vehicle moves forward with driving force F and experiences resistance R. The net real force is F - R, causing acceleration a. By D'Alembert, we write F - R - ma =
7. A 10 N force rightward at (0,0) and a 5 N·m clockwise couple at (0,0) act on a body. Find an equivalent single force at (0,1) m.
We want only a force at (0,1) m that gives the same total force and moment about any point as the original. The total force is already 10 N rightward. So at (0,1) we place 10 N rightward. Now check the moment about (0,0). Original moment about (0,0) is just the couple: -5 N·m (clockwise). The new force at (0,1) creates a moment about (0,0): r × F. r is (0,1) m, F is (10,0) N. Cross product gives (0 × 0 - 0 × 0, 0 × 10 - 0 × 0, 0 × 0 - 1 × 10) = (0,0,-10) N·m = 10 N·m clockwise. That is not -5; we need to add a couple to adjust. So just a single force at (0,1) cannot match unless we also add a moment. To get an equivalent single force, we must place the force at a point where its moment equals the original total moment. For that, we need a point (0, y) such that r × F = -5. r = (0, y), F = (10,0), so r × F = (0,0, -10y) = -5 => -10y = -5 => y = 0.5 m. So the equivalent single force is 10 N rightward at (0, 0.5) m. At (0,1) m, we cannot have only a force; we need a force and a couple.
8. Given a string with linear mass density $\mu$ and tension $T$, derive the Lagrangian density and then the wave equation.
For the string, the kinetic energy density is $\frac{1}{2}\mu (\partial y/\partial t)^2$ and potential energy density is $\frac{1}{2}T (\partial y/\partial x)^2$. So the Lagrangian density is $\mathcal{L} = \frac{1}{2}\mu (\partial_t y)^2 - \frac{1}{2}T (\partial_x y)^2$, where we write $\partial_t = \partial/\partial t$, $\partial_x = \partial/\partial x$. Apply the Euler–Lagrange equation for fields: $\partial_t (\partial \mathcal{L}/\partial (\partial_t y)) + \partial_x (\partial \mathcal{L}/\partial (\partial_x y)) - \partial \mathcal{L}/\partial y = 0$. Compute: $\partial \mathcal{L}/\partial (\partial_t y) = \mu \partial_t y$, so $\partial_t (\mu \partial_t y) = \mu \partial_t^2 y$. Similarly, $\partial_x( -T \partial_x y) = -T \partial_x^2 y$, and $\partial \mathcal{L}/\partial y = 0$. Thus, $\mu \partial_t^2 y - T \partial_x^2 y = 0$, or the wave equation $\partial_t^2 y = (T/\mu) \partial_x^2 y$.
9. Why are square threads often preferred for power screws over V-threads in terms of efficiency?
Square threads are more efficient because they waste less energy to friction. In a V-thread, the sloping flanks create a wedging action. The normal force between threads is larger than the axial load, because the force is split into components. This larger normal force increases friction for the same load. In a square thread, the flanks are parallel to the axis, so the normal force equals the axial load. Thus, friction is lower. Another reason is that V-threads often have a larger contact area, which does not directly increase friction force (friction depends on normal force, not area), but it can cause more wear and uneven loading. Square threads are easier to machine and give a higher mechanical advantage for moving loads. However, they are harder to make with split-nuts for engaging and disengaging, so sometimes modified trapezoidal threads are used.
10. Prove that the moment of a force about a point equals the sum of moments of its components using a simple example.
Consider a force F at point (x,y) making an angle with the horizontal. Let its components be Fx horizontally and Fy vertically. The moment of F about the origin is F times the perpendicular distance d. By geometry, d = |x sinθ - y cosθ|? Actually, moment = x Fy - y Fx (cross product). Now, moment of Fx about origin: Fx has lever arm y (vertical distance), and it causes rotation around origin: moment = -Fy * x? Let's keep sign. If Fx is along x, its moment = -y * Fx? Actually, position vector (x,y) cross (Fx,0) = x*0 - y*Fx = -y Fx. Moment of Fy: (x,y) cross (0, Fy) = x*Fy - y*0 = x Fy. So total = x Fy - y Fx, which is exactly the moment of F. A numerical example: F=10 N at (3,4) at angle? Suppose F points exactly such that components are 6 and 8. Moment = 3*8 - 4*6 = 24-24=0. So moment zero. Thus, the sum of component moments equals the total moment.
11. A block sits on a horizontal rough surface. Explain how you can use the angle of friction to find the coefficient of friction.
First, apply a horizontal force to the block and slowly increase it until the block just starts to move. At that moment, the horizontal force equals the limiting friction force. You also know the normal force is the weight of the block. The ratio of limiting friction to normal force gives the coefficient of static friction. To use the angle of friction, instead push with a force at an angle. Tilt the applied force until the block just starts to slide. When the applied force makes an angle with the surface normal that equals the angle of friction, the block is on the verge of motion. Measure this angle with a protractor or calculate it from forces. The tangent of that angle is the coefficient of friction. This method works because the angle of friction is the angle whose tangent is the friction coefficient.
12. How does rolling resistance differ from sliding friction?
Sliding friction happens when two surfaces rub directly against each other, like a box being dragged. It comes from surface roughness interlocking and sometimes slight sticking. The force is proportional to the normal load times a coefficient that is usually between 0.2 and 0.8 for dry surfaces. Rolling resistance is different: it happens because the rolling object and the surface are not perfectly stiff; they dent. The main force comes from the energy needed to keep making this dent as the object rolls. The coefficient of rolling friction is much smaller, often 0.001 to 0.05 for smooth wheels on hard surfaces. Also, sliding friction does not depend on speed much, while rolling resistance can increase slightly with speed. Using wheels or ball bearings changes sliding to rolling, greatly reducing friction.