find the tension and acceleration of bodies in which one body is moving vertically downward and other move horizontal on a table over a frictionless pulley
Answers
Answer:
Explanation:
Two bodies A & B of masses m1 and m2 are attached to the ends of a string, which passes over a frictionless pulley as shown in the figure. The body "A" moves vertically downward with an acceleration equal to "a" and the body "B" moves on a smooth horizontal plane towards the pulley with the same acceleration.
Consider the motion of body A
Forces acting on the body A
• Weight of the body: w1 = m1g
• Tension in the string = T
The net force acting on the body A
m1g – T
According to Newton’s 2nd law of motion, the resultant force acting on it is equal to m1a
Therefore,
m1g – T = m1a --------- (i)
Now Consider the motion of body B
There are three forces acting on it.
• The tension (T) in the string, which acts horizontally towards the pulley.
• Its weight w2 = m2g which acts vertically downward.
• Reaction of the surface (R) on the body which acts vertically upward.
As there is no motion of body "B" in the vertical direction. Therefore weight and normal reaction cancel each other
Thus the net horizontal force acting upon body B is T
Applying Newton’s 2nd law of motion, we get
T = m2a --------(ii)
Putting the value of T in equation (i), we get
m1g – m2a = m1a
m1g = m1a + m2a
m1g = a (m1 + m2)
Putting the value of "a" in equation (ii), we get
T = m2a --------(ii)
T = m2
Answer:
Explanation:
Two bodies A & B of masses m1 and m2 are attached to the ends of a string, which passes over a frictionless pulley as shown in the figure. The body "A" moves vertically downward with an acceleration equal to "a" and the body "B" moves on a smooth horizontal plane towards the pulley with the same acceleration.
Consider the motion of body A
Forces acting on the body A
• Weight of the body: w1 = m1g
• Tension in the string = T
The net force acting on the body A
m1g – T
According to Newton’s 2nd law of motion, the resultant force acting on it is equal to m1a
Therefore,
m1g – T = m1a --------- (i)
Now Consider the motion of body B
There are three forces acting on it.
• The tension (T) in the string, which acts horizontally towards the pulley.
• Its weight w2 = m2g which acts vertically downward.
• Reaction of the surface (R) on the body which acts vertically upward.
As there is no motion of body "B" in the vertical direction. Therefore weight and normal reaction cancel each other
Thus the net horizontal force acting upon body B is T
Applying Newton’s 2nd law of motion, we get
T = m2a --------(ii)
Putting the value of T in equation (i), we get
m1g – m2a = m1a
m1g = m1a + m2a
m1g = a (m1 + m2)
Putting the value of "a" in equation (ii), we get
T = m2a --------(ii)
T = m2