given answer may be wrong. The two blocks A and B are connected by a light strin
ID: 1972059 • Letter: G
Question
given answer may be wrong.
The two blocks A and B are connected by a light string that passes over a frictionless pulleys, as shown below. The mass of block A is 10.0 kg, and the mass of blocks B is 25.0 kg. The coefficient of friction between block and incline surface is 0.08. Draw free-body force diagram of block A and block B. Create the force table for block A and block B based on your reference flame. Write down the Newton's component equations for A and B. Find the acceleration of block A and the tension force of the string.Explanation / Answer
1) The forces acting on A : T,mgsin45,N,mgcos45; [m = mass of A ]
T-tension in the rope,
mgsin45 is component of its weight acting downwards along the incline,
mgcos45 is component of its weight acting downward along the normal direction to the plane,
N is normal reaction at the surface contact of A and the incline acting upwards on A and normal to the plane.
The forces acting on B : T,Mg [M - mass of B]
T - tension in the rope acting upward along the rope
Mg - weight acting along downward direction on B
2) & 3) Net forces acting on A :
N = mgcos45 ; T = mgsin45
Net forces acting on B :
T = Mg ;
The above equations are for static equilibrium of the system
4) In motion, an extra force f(kinetic friction) comes into the picture and this acts opposite to the motion of a block along its contact surface
Let A is going upwards along the incline and B is moving downward .
The equation of motions are :
for A :
T - mgsin45 - f = ma ;
f = N ; [-coefficient of friction]
N = mgcos45 = mg/2 ;
=> T - N - mg/2 = ma ;
=> T - 0.08(mg/2) -mg/2 = ma ;
=> T -7.637g = 10a ; -----------------------------------(i)
for B :
Mg - T = Ma ; ------------------------------------------(ii)
adding (i),(ii)
=> Mg - 7.637g = 35a
=> a = 4.86 m/s2 [g = 9.8m/s2]
and T = M(g-a) from (ii)
=> T = 25(9.8-4.86) = 123.5 N
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