Question

Review. A block of mass m₁=2.00Kg and a block of mass m₂=6.00Kg are connected by a mass less string over a pulley in the shape of a solid disk having radius R=0.250m and mass m=10.0kg.The fixed, wedge shaped ramp makes an angle of u=30.08as shown in Figure

The coefficient of kinetic friction is 0.360 for both blocks.

(a) Draw force diagrams of both blocks and of the pulley. Determine

(b) The acceleration of the two blocks and

(c) The tensions in the string on both sides of the pulley

Short answer

a) The force diagram of both blocks and of the pulley is given below .

b) The acceleration of the two blocks is $0.422m/s^2$.

c) The tensions in the string on both sides of the pulley.

$\begin{array}{l}{T}_1=6.72N\\ T_2=8.83N\end{array}$

Step-by-step solution

Given data.

The given data is,

The mass of the block is, 2.00Kg,

The another block of mass is m₂=6.00Kg,

A solid disk having radius R=0.250m,

A solid disk having mass, M=10.0Kg

And has an angle of The coefficient of kinetic friction is 0.360.

Force diagrams:

a)

The force diagram of both blocks and of the pulley is given below.

Acceleration of the two blocks.

b)

The two blocks will move with the same acceleration a

• Newtons second law for block gives :

$\begin{array}{l}{T}_1-f_{k1}=m_{1}a\\ T_1=\mu m_{1}g+m_{1}a\end{array}$

• Newtons second law for block m2 gives :

$\begin{array}{l}-T_2-f_{k2}+m_{2}gsin\theta =m_{2}a\\ T_2=m_{2}gsin\theta -\mu m_{2}gcos\theta -m_{2}a\end{array}$

• For the pulley (considered a disk with moment of inertia MR2 2 ):

$\begin{array}{l}\begin{array}{l}\tau net=I\alpha \\ T_{2}R-T_{1}R=I\frac{a}{R}\end{array}\\ a=\frac{R^2}{\frac{M{R}^2}{2}}(T_2-T_1)\end{array}$

Rearranging should give:

$\begin{array}{l}a=\frac{2}{M}(m_{2}gsin\theta -\mu m_{2}gcos\theta -\mu m_{1}g)-\frac{2}{M}(m_1+m_2)a\\ \\ =\frac{2g(m_{2}sin\theta -\mu m_{2}cos\theta -\mu m_1)}{M+2m_1+2m_2}\\ \\ =\frac{2\times 9.81(6.20sin30-0.360\times 1.70\times cos30-0.360\times 1.70)}{10+2\times 1.70+2\times 6.20}m/s^2\\ \\ =0.422m/s^2\end{array}$

Tensions in the string on both sides of the pulley.

c)

To the LEFT of the pulley, the string tension is T1:

$\begin{array}{l}{T}_1=\mu m_{1}g+m_{1}a\\ =1.70(0.360\times 9.81+0.422)N\\ =6.72N\end{array}$

To the RIGHT of the pulley, the string tension is T2:

$\begin{array}{l}{T}_2=m_{2}gsin\theta -\mu m_{2}gcos\theta -m_{2}a\\ =6.20(9.81sin30-0.360\times 9.81\times cos30-0.422)N\\ =8.83N\end{array}$

Hence, the tensions in the string on both sides of the pulley are 8.83N.