Refer to the diagram a side view The pulley system here is c

Refer to the diagram (a side view). The pulley system here is completely ideal (all the wheels and wires have negligible mass). At time t = 0 s, a steady force F is applied to wire 1, as shown, to move a sled (mass m_s) up a frictionless ramp. (Assume g = 9.80 m/s^2.) At all times, the sled\'s top surface remains level. On that surface is a box of mass m_B. Friction (mu_S, mu_k) is available between the sled surface and the box. The data: F = 428N mu_S = 0.517 mu_K = 0.398 m_B = 12.6 kg m_S = 65.9 kg theta = 34.7 degree a. Find v(t), the velocity of the sled as a function of time, assuming that it starts at rest at time t = 0. You may use any coordinate system you choose for your calculations, but express your final result as a vector using the coordinate system indicated. b. Calculate the friction force (magnitude and direction) acting on the box while F is being applied. Again, express your final result as a vector using the x-y coordinate system indicated. c. Calculate the box\'s altitude (vertical position) change during the time interval 1.00 lessthanorequalto t lessthanorequalto 3.00 s.

Solution

solution:

1) here whensteady force F is applied to wire 1 then tension in wire is

T=F

this tension acting in two side of pully sum together and act down ward at center of pully hence force at center

F\'=2F

2) hence this F\' is act as tension in second wire and over the pully it is transfer to sled,hence forces acting on sled are

1) tension acting through wire 2=F\'=2F=856 N

2) taotal weight vertical down ward=(ms+mb)g=770.085 N

3) friction force in negative x direction due to box at top surface=Fr=.915*123.606=113.099 N

3) as motion takes place along the ramp hence forces along the ramp are producing accelaration a,hence

2F-Wsin34.7-frcos34.7=(ms+mb)a

856-770.085sin34.7-113.099cos34.7=(78.5)a

a=4.1353 m/s2

where velocity is given byalong ramp as

v=vo+at      if vo=0 m/s

v=at=4.1353 t

v=Vxi+Vyj=(3.39*t)i+(2.3441*t)j

4)here force on box is frictional force

Fr=.915*mb*9.81=113.099 N

in x,y coordinate as

Fr=fxi+fyj

as angle=0 for horizontal force

Fr=fxi=113.099i+0j

5) where altitude of box is depend on sled motion

d=v*t=4.1353 t^2

hence distance move on ramp during 1 to 3 sec

d=33.0824 m

where altitude h=dsin34.7=18.83 m