The polar coordinate of point a of the crane are given as functions of time in s

Question

The polar coordinate of point a of the crane are given as functions of time in seconds by r = l - cos(Pi/3. t) m and Theta = 2 - 2 sin (Pi/4. T) rad. Determine the (1) velocity and (2) acceleration of pant a m terms of polar coordinate) at t = 2 s. The mass of crate is 10 kg and the angle of surface is 20 degrees and the crate at rest is about to move What is the minimum static friction coefficient between the crate and surface to make the a ate stationary? If the static friction coefficient is now 0.1 and you can change the angle of surface from 20 deg, to what angle should it be changed to make the crate stationary? When the kinetic friction coefficient is 0.05, find the acceleration. Using the answer of 3) find the velocity of the crate at 2 seconds. Using the answer of 3) find the velocity of the crate when it has moved 5 m from the ordinal position. how long would it take that the crate travels 10 m from the original position. When a is zero and v is constant, s = s_o + vt When a is constant, s = s_o + vt + 1/2at^2, v = v_o + at, v^2 + 2a(s - s_o), Other definition a = dv/ct, v = ds/dt, a = v dv/ds Velocity and acceleration m tangential normal coordinates v Bar = v_t e Bar_t, a Bar = a_t e Bar_t + a_n e Bar_n, = a_t e Bar_t + v_t w e Bar_n Where v Bar_t = R w Bar, a_t = dv_t/dt, w = dTheta/dt Position, velocity and Acceleration in polar coordinates r Bar = r e Bar_r, v Bar = v_r e Bar_r + v_Theta e Bar_Theta, = v_r e Bar_r + r w e Bar_Theta, a Bar = (a_r - rw^2) e Bar_r + (2v_r w + ra)e Bar_Theta Where a_r = dv_r/dt, v_r = dr/dt, w = dTheta/dt, a = dw/dt Friction force: f = Mu N where N is normal force.

Explanation / Answer

Drawing the FBD of the crate properly one gets a down the incline force of Fd = m*g*sin and an up the incline force of Fu = *N = *m*g*cos , where is the coefficient of friction.

1. Now for the crate to be stationary one needs:

Fu = Fd   or   m*g*sin = *m*g*cos or = tan = tan20 = 0.364

2. Again with = 0.1, the equilibrium condition requires that:

= tan where    is the new inclination of the surface.

Therefore     = tan-1(0.1) = 5.710

3. Now with = 200   and k = 0.05 we get the kinematical equation using Newton's second law as:

Fd - Fu = m*a or m*g*sin - *m*g*cos = m*a or a = g*(sin - *cos)

So a = 9.8*(sin20 - 0.05*cos20) = 2.89 m/s2

4. using second equation of motion we get velocity after t = 2 s:

V = 0 + 2.89*2 = 5.78 m/s

5. Using third equation of motion we get velocity for x = 5m as:

v2 = 0 + 2*2.89*5 = 28.9 m/s

6. Time taken for the crate to travel 10m can be calculated using second equation of motion as:

10 = 0*t + 0.5*2.89*t2   or t = (20/2.89)0.5  = 2.63 s

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