Newton's Laws 7

Question

An object with mass m1 = 5.00 kg, rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 9.0 kg, as shown in Figure P4.30. Find the acceleration of each object and the tension in the cable.

 

mass m1________m/s2

 

mass m2________m/s2

 

tension__________N

Answer 1

Let us analyze the forces on each mass in turn.

 

Mass m₂ has an upward tension T pulling up and a weight m₂g pulling downward.  Then the net downward force on m₂ is

 

m₂a₂ = (m₂g - T) ⇒ a₂ =  (m₂g - T)/m₂

 

The pulley merely changes the direction of T to a horizontal direction.  Since there is no friction,  

 

m₁a₁ = T

 

Since m₁ and m₂ are connected, the horizontal acceleration of m₁ is the same in magnitude as the vertical downward acceleration of m₂.  The two masses have the same acceleration but in different directions.  Call that acceleration a.

 

|a₂| = |a₁| = a

 

Then

 

(i)  T = m₁a

(ii) a = (m₂g - T)/m₂

 

Solve these two equations above for a and T.  Substitute T from (i) into (ii) and get

 

a = (m₂g - m₁a)/m₂

 

Solve for a and get

 

a = [m₂/(m₁ + m₂)] g = [9.00/(5.00 + 9.00)]*9.81 m/s² ≅ 6.306 m/s²

 

For m₁, a is toward the pulley; for m₂, a is down.

 

T = m₁a = 5.00*6.306 N = 31.53 N

 

On m₁, T pulls toward the pulley; on m₂, T pulls up.

 

 

 

 

 

Answer 2

Since the two masses are attached to the same cable, the cable exerts the same force on both masses due to Newton's Third Law.

 

The hanging masses experiences a downward tension, while the mass on the table experience a horizontal tension.

 

T = -m₂g

 

F = m₁a

 

 

Therefore,

 

T = F

 

-m₂g = m₁a

 

 

g = 9.81 m/s²   ------>  acceleration of mass 2

 

 

You can get the tension by plugging in  g=9.81  and  m₂=9.

 

To get the acceleration of mass 1, calculate      a₁ = T / m₁