In the system shown, m = 4 kg, 0 = 40°, µ¸ = 0.5 and µk = 0.3 and the wheels are frictionless.(a) Determine the maximum value of force P for the system to have static equilibrium. Call thisforce Pmax (Pmax=105.7 N)(b) Determine the force P for the system to have dynamic equilibrium, that is, moves with constantvelocity, while moving up the incline. (P = 93.7 N)=2Pmax, determine the accel-(c) If the system is released from rest with the cable taught and Peration of the bodies and the tension of the cable. (a = 9.81 m/s², T = 147 N)μς, μκ <2mᎾmP

Answered: Image /qna-images/answer/425b2172-2ad5-44dd-b891-6baa94c4421d.jpg

Answered: In the system shown, m = 4 kg, 0 = 40°,…

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter2: Right Triangle Trigonometry

Section: Chapter Questions

Problem 28CT: Force Tyler and his cousin Kelly have attached a rope to the branch of a tree and tied a board to...

See similar textbooks

Similar questions

Force Tyler and his cousin Kelly have attached a rope to the branch of a tree and tied a board to the other end to form a swing. Tyler sits on the board while his cousin pushes him through an angle of 25.5 and holds him there. If Tyler weighs 95.5 pounds, find the magnitude of the force Kelly must push with horizontally to keep Tyler in static equilibrium. See Figure 1. Figure 1
Brazilian soccer star Marta has a penalty kick in the quarter-final match. She kicks the soccer ball from ground level with the (x, y)-coordinates (76, 21) on the soccer field shown in the figure and with initial velocity vo = 8i - 4j+23k ft/s. Assume an acceleration of 32 ft/s² due to gravity and that the goal net has a height of 8 ft and a total width of 24 ft. 105 ft 105 ft ri = rj = 165 ft rk = e 165 ft Determine the position function that gives the position of the ball t seconds after it is hit. (Use symbolic notation and fractions where needed.) r(t) = r(t)i + r(t)j +rk(t)k 12 12 X
A basket of flowers of mass 3 kg is placed on a flat grassy slope that makes an angle θ with the horizontal. The coefficient of static friction between the basket and the slope is 0.45 and the basket is on the point of slipping down the slope. Model the basket of flowers as a particle and the grassy slope as a plane. Take the magnitude of the acceleration due to gravity, g, to be 9.8 m s−2 Express the forces in component form, in terms of θ and unknown magnitudes where appropriate. Write down the equilibrium condition for the basket and hence show that tan θ = 0.45. Determine the angle, in degrees, that the slope makes with the horizontal.
A seasoned parachutist went for a skydiving trip where he performed freefall before deploying the parachute. According to Newton's Second Law of Motion, there are two forcës acting on the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa) as shown in Figure 1. Fa = -cv ITM EUTM FUTM * UTM TM Fg= -mg x(t) UTM UT UTM /IM LTM UTM UTM TUIM UTM F UT GROUND Figure 1: Force acting on body of free-fall where x(t) is the position of the parachutist from the ground at given time, t is the time of fall calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration, v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the position is given by the equations below: EUTM PUT v(t) = mg -et/m – 1) (Eq. 1.1) x(t) = x(0) – Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the critical position to deploy the parachutes is at 762 m from the ground…
Which of the following statements is true for regular, isotropic beams: Maximum bending or flexural stress occurs at the point in a cross-section that is furthest from the neutral axis. Flexural stress varies linearly along the cross-section of the beam. (A) B) (D) (E) F Maximum horizontal shearing stress occurs at the neutral axis. Horizontal shearing stress varies "parabolically" along the cross-section of the beam. Maximum bending or flexural stress occurs at the neutral axis. Flexural stress varies linearly along the cross-section of the beam. Maximum horizontal shearing stress occurs at the neutral axis. Horizontal shearing stress varies "parabolically" along the cross-section of the beam. Maximum bending or flexural stress occurs at the neutral axis. Flexural stress varies linearly along the cross-section of the beam. Maximum horizontal shearing stress occurs at the point in a cross-section that is furthest from the neutral axis. Horizontal shearing stress varies "parabolically"…
vn] An equation of the motion of the form s = 80t – 16r? represents an object thrown upward with initial velocity of 80 ft/sec. a. Find the maximum height. the velocity b. Find acceleration of the object when height is 96 feet. and
(3) Compute the couple by the two parallel forces shown for F = 65 lb. Ans: M=2300 lb-ft CCW 20% F y, ft A (-10, 14) C (-12, -12) D (20, 0) x, ft B (20,-7) F 20°
Assume that an object moves on a trajectory according to the equations x(t)=t+cost, y(t)=t−sint. Find the magnitude of the acceleration vector.
Find the maximum rate of change of f(x, y) Maximum rate of change: 2/5 In(x? + y?) at the point (-4, -3) and the direction in which it occurs. Direction (unit vector) in which it occurs: -8/25 -6/25 出)
An ant walks due east at a constant speed of 2 mi/hr on a sheet of paper that rests on a table. Suddenly, the sheet of paper starts moving due southeast at v2 mi/hr. The following is a correct solution for finding the resultant speed and direction of the ant relative to the table. Set up a coordinate system so that the positive y-axis corresponds to north and the positive x-axis corresponds to east. Then the velocity vector for the ant relative to the paper is given by (2,0). The velocity vector for the paper relative to the table is given by (v2 cos (-4), v2 sin (-5))= (1, –1). Therefore, the resultant velocity vector is given by (2,0) + (1, –1) = (3, –1). The speed of the ant relative to the table is v32 + 1² = v10 mi/hr and the direction the ant is traveling relative to the table is found by (금) using tan 0 = -1 where 0 represents the angle made with the positive r-axis. In this case 0 = arctan 3 3
Parametrize y = -x2 + 4 in terms of u and v %3D
Given the position vector r(t) = the equations of the velocitiy vector, acceleration vector and, speed are v(t) = , a(t) = , s(t) =/9sin?t+16 cos?t v(t) = , a(t)= , s(t) = /9sin?t+16cos?t v(t) = , a(t) = ,s(t) =5 v(t) = , a(t) = , s(t) =5

SEE MORE QUESTIONS

Recommended textbooks for you

Trigonometry (MindTap Course List)

Trigonometry

ISBN:

9781305652224

Author:

Charles P. McKeague, Mark D. Turner

Publisher:

Cengage Learning

Trigonometry (MindTap Course List)

Trigonometry

ISBN:

9781337278461

Author:

Ron Larson

Publisher:

Cengage Learning

Trigonometry (MindTap Course List)

Trigonometry

ISBN:

9781305652224

Author:

Charles P. McKeague, Mark D. Turner

Publisher:

Cengage Learning

Trigonometry (MindTap Course List)

Trigonometry

ISBN:

9781337278461

Author:

Ron Larson

Publisher:

Cengage Learning

SEE MORE TEXTBOOKS