Respuesta :
Answer:
v = 0.384 m/s
Explanation:
The complete problem is as follows:
" The 45-kg crate is pulled up the inclined surface by the winch. The coefficient of kinetic friction between the crate and the surface is µk = 0.4.
The moment of inertia of the drum on which the cable is being wound is IA = 4 kg-m2
.The crate starts from rest, and the motor exerts a constant couple M = 50 N-m on the drum. Use the principle of work and energy to
determine the magnitude of the velocity of the crate when it has moved 1 m "
Given:
- The mass of the crate m = 45 kg
- Coefficient of kinetic friction µk = 0.4
- Moment of inertia of the drum IA = 4 kg-m2
- Motor constant couple M = 50 N-m
- Motor Couple moment arm R = 0.15 m
- (The missing figure is attached)
Find:
Use the principle of work and energy to
determine the magnitude of the velocity of the crate when it has moved 1 m
Solution:
Develop a FBD of the system consisting of Motor, rope and crate. (Attachment)
The motor generates a couple M which induces a tension in the string depending on the amount of rotation proportional to the distance moved by the block.
The crate experiences 3 forces: Normal contact force (N) exerted by the inclined surface on the crate. The Weight (W) of the crate acting vertically downwards and Friction Force (Ff = uk*N) along the surface contact parallel to slope while opposing uphill motion. ( See Attachment )
- Apply Equilibrium condition on the crate perpendicular to the slope:
N - m*g*cos(20) = 0
N = m*g*cos(20)
- For motion of the crate as it moves 1 m along the surface the work done is given by U_12:
U_12 = Work done by motor couple - Work done against gravity - Work done by friction
U_12 = M*θ - m*g*sin(20) - uk*N
Where, θ is the angle of rotation of motor through distance of 1 m = 1 / R.
U_12 = M/R - [m*g*sin(20) - uk*m*g*cos(20)]*(1)
- The kinetic energy of system at initial condition is zero ( T_1 = 0 ). The final state Kinetic energy are as follows T_2:
T_2 = 0.5*m*v^2 + 0.5*IA*w^2
Where,
w = v / R ..... Angular speed at which the motor disc rotates.(No slip)
T_2 = 0.5*m*v^2 + 0.5*IA*v^2 / R^2
T_2 = v^2*( 0.5*m + 0.5*IA / R^2 )
- The Total Energy balance would be:
T_1 + U_12 = T_2
0 + M/R - m*g*sin(20) - uk*m*g*cos(20) = v^2*( 0.5*m + 0.5*IA / R^2 )
v^2 =[ M/R - m*g*sin(20) - uk*m*g*cos(20) ] / [ ( 0.5*m + 0.5*IA / R^2 )]
- Plug in values:
v^2 =[ 50/0.15 - 45*9.81*sin(20) - 0.4*45*9.81*cos(20) ] / [ ( 0.5*45 + 0.5*4 / 0.15^2 )]
v = sqrt (0.147456)
v = 0.384 m/s
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