The figure below shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from a point straight above its mast. At the particular speed of the sailboat, the water exerts aF-178-N drag force on its hull and 0-49.0", For each of the situations (a) and (b) described below, write two component equations representing Newton's second law. Then solve the equations forP (the force exerted by the wind on the sail) and for in the force exerted by the water on the keel).(a) Choose the x direction as east and the y direction as northA-1546✔N✔NP235.0(b) Now choose the I direction as 0-49.0 north of east and the y direction as 0-49.0 west of northA-156.7N2041xa number differs from the correct answer by more than 10%. Docble check your calculations. NP=

The figure below shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from a point straight above its mast. At the particular speed of the sailboat, the water exert -178-N drag force on its hull and 0-49.0. For each of the situations (a) and (b) described below, write two component equations representing Newton's second law. Then solve the equations t P (the force exerted by the wind on the sail) and for n (the force exerted by the water on the keel). (a) Choose the x direction as east and the y direction as north A-1546 P235.8 (b) Now choose the direction as 8-49.0 north of east and the y direction as 0-490 west of north 1567 N 204

The figure below shows the horizontal forces acting on a sailboat moving north at constant velocity, seen from a point straight above its mast. At the particular speed of the sailboat, the water exert -178-N drag force on its hull and 0-49.0. For each of the situations (a) and (b) described below, write two component equations representing Newton's second law. Then solve the equations t P (the force exerted by the wind on the sail) and for n (the force exerted by the water on the keel). (a) Choose the x direction as east and the y direction as north A-1546 P235.8 (b) Now choose the direction as 8-49.0 north of east and the y direction as 0-490 west of north 1567 N 204

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter2: Newtonian Mechanics-single Particle

Section: Chapter Questions

Problem 2.9P: Consider a projectile fired vertically in a constant gravitauonal field. For the same initial...

See similar textbooks

Similar questions

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, taken as the xy plane. The 4.00-kg puck has a velocity of 3.00im/s at one instant. Eight seconds later, its velocity is (8i+10j)m/s. Assuming the rocket engine exerts a constant horizontal force, find (a) the components of the force and (b) its magnitude.
A particle of mass m has speed υ = α/x, where x is its displacement. Find the force F(x) responsible.
Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00104kg . The thrust of its engines is 3.00104N . (a) Calculate the module’s magnitude of acceleration in a vertical takeoff from the Moon. (b) Could it lift off from Earth? If not, why not? If it could, calculate the magnitude of its acceleration.
Consider a projectile fired vertically in a constant gravitauonal field. For the same initial velocities, compare the times required the projectile to reach its maximum height (a) for zero resisting force, (b) for a resisting force proportional to the instantaneous velocity of the projectile.
A 6 kg crate is pulled by a force of 185 N at an angle of 18^0 from the horizontal. A constant frictional force of 47 N resisted the crate. Calculate the net force experiences by the crate
Write Newton's laws for a static system (1) EF, = mg sin 0 – µ̟n = 0 in component form. The gravity force has two components. (2) EF, = n - mg cos 0 = 0 Rearrange Equation (2) to get an n = mg cos e expression for the normal force n. Substitute the expression for n into EF, = mg sin 0 – µ̟mgcos 0 = 0 → tan 0 =µ, Equation (1) and solve for tan 0. Apply the inverse tangent function to tan 0 = 0.350 → 0 = tan 1 (0.350) = 19.3° get the answer. LEARN MORE REMARKS It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully. QUESTION A larger static friction constant would result in a: (Select all that apply.) O larger component of normal force at the maximum angle. O larger component of gravitational force along the ramp at the maximum angle. smaller component of gravitational force along the ramp…
A force of 600 N is applied on a block of mass M to pull it at a constant velocity up the incline at an angle of 30degrees with the horizontal. The coefficient of kinetic friction between the surface of the incline and the block is 0.5. Calculate the mass of the block.
A crate of mass 10 KG eat in Stationæry on an incline surface of angle equal to 30° and upward force of 50 N is applied to the create parallel to the incline but the crate remain stationary despite the applied force determine the magnitude in the direction of the frictional force acting on the crate
Two blocks our positioned on surfaces, Each inclined at the same angle at 41.1° with respect to the horizontal. The blocks are connected by a rope which rests on a frictionless pulley at the top of the inclines a Sean, so the blocks can slide together. The mass of the black block is 7.35 kg, and the coefficient of kinetic friction for both blocks and inclines is 0.290. Assume static friction has been overcome and that everything can slide. What must be the mass of the white block if both blocks are to slide to the right at an acceleration of 1.5 m/s^2?
a block of mass m is initially traveling upward along a ramp at speed v. The coefficient of kinetic friction between the block and the ramp is Mk. (a) Using Newton's second law, solve for the normal force on the block due to the ramp. (b) Using Newton's second law solve for the acceleration of the block.
my and Tadiany 28. A block of of mass m, = 4=Ikg another mass m, =2 kg, are placed together (see figure) on an inclined plane with angle of inclination 9. Various values of A are given in List-I. The coefficient of friction between the block m, and the plane is always zero The.coefficient of static and dynamic friction between the blocķ m, and the plane are equal to u =0.3: 1Tn List-T expressiöns for the frictioni on block m, are given. Match the coirect expression of the friction in Lişt-II with the angles given in List-I, and the choose the correct option. The acceleration due to gravity is denoted by g. [Useful inforınation : tan tan List-I List-I (Р) 0 — 5° (1) m2g sine (Q) 0 = 10° (2) (m +m2)gsin 0 (R) 0 = 15° (3) um,g cos 0 (S) 0 = 20°:? (4) µ(m, + m2)g cose (а) Р-1, Q-1, R-1, S-3 (с) Р-2, Q-2, R-2, S-4 (b) Р-2, Q-2, R-2, S-3 (d) Р-2, Q-2, R-3, S-3
Enter an expression for the magnitude of the force of the bottom-most parcel on the center parcel valid for any value of the vertical acceleration, ay, using only the parameters provided in the palette