Questions: Two blocks connected by a rope of negligible mass are being dragged by a force at a 32° angle above horizontal, so that the blocks are already moving to the right. Suppose F=75.0 N, m1=10.0 kg, m2=16.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.150. Use g=9.8 m/s^2. m1 m2 F (b) Determine the acceleration of the system. 1.21 m/s^2 (c) Determine the tension T in the rope. 13.91 × N

Two blocks connected by a rope of negligible mass are being dragged by a force at a 32° angle above horizontal, so that the blocks are already moving to the right. Suppose F=75.0 N, m1=10.0 kg, m2=16.0 kg, and the coefficient of kinetic friction between each block and the surface is 0.150. Use g=9.8 m/s^2.
m1 m2 F
(b) Determine the acceleration of the system.
1.21 m/s^2
(c) Determine the tension T in the rope.
13.91 × N

Transcript text: Two blocks connected by a rope of negligible mass are being dragged by a force at a $32^{\circ}$ angle above horizontal (see figure below), so that the blocks are already moving to the right. Suppose $F=75.0 \mathrm{~N}, m_{1}=10.0 \mathrm{~kg}, m_{2}=16.0 \mathrm{~kg}$, and the coefficient of kinetic friction between each block and the surface is 0.150 . Use $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$. $m_{1}$ $\square$ $m_{2}$ $\square$ $\overrightarrow{\mathrm{F}}$ (b) Determine the acceleration of the system. $\qquad$ 1.21 $\mathrm{m} / \mathrm{s}^{2}$ (c) Determine the tension $T$ in the rope. 13.91 $\times N$

Solution

Solution Steps

Step 1: Identify the Forces Acting on the System

We need to identify all the forces acting on the two-block system. The forces include:

The applied force $ F $ at an angle $ \theta = 32^\circ $
The gravitational force on each block $ m_1 $ and $ m_2 $
The normal force on each block
The kinetic friction force on each block

Step 2: Calculate the Normal Force

The normal force $ N $ for each block can be found by considering the vertical forces. For block $ m_1 $: \[ N_1 = m_1 g - F \sin(32^\circ) \] For block $ m_2 $: \[ N_2 = m_2 g \]

Step 3: Calculate the Frictional Force

The kinetic friction force $ f_k $ for each block is given by: \[ f_{k1} = \mu_k N_1 \] \[ f_{k2} = \mu_k N_2 \]

Step 4: Calculate the Net Force in the Horizontal Direction

The net force $ F_{\text{net}} $ acting on the system is the applied force minus the frictional forces: \[ F_{\text{net}} = F \cos(32^\circ) - f_{k1} - f_{k2} \]

Step 5: Calculate the Acceleration of the System

Using Newton's second law, the acceleration $ a $ of the system is: \[ a = \frac{F_{\text{net}}}{m_1 + m_2} \]

Step 6: Calculate the Tension in the Rope

To find the tension $ T $ in the rope, we consider the forces acting on one of the blocks (e.g., $ m_1 $): \[ T = m_1 a + f_{k1} \]

Final Answer

Let's plug in the numbers and calculate the values:

Normal force for $ m_1 $: \[ N_1 = m_1 g - F \sin(32^\circ) = 10.0 \times 9.8 - 75.0 \sin(32^\circ) \approx 98.0 - 39.8 = 58.2 \, \text{N} \]
Normal force for $ m_2 $: \[ N_2 = m_2 g = 16.0 \times 9.8 = 156.8 \, \text{N} \]
Frictional forces: \[ f_{k1} = \mu_k N_1 = 0.150 \times 58.2 \approx 8.73 \, \text{N} \] \[ f_{k2} = \mu_k N_2 = 0.150 \times 156.8 \approx 23.52 \, \text{N} \]
Net force: \[ F_{\text{net}} = 75.0 \cos(32^\circ) - 8.73 - 23.52 \approx 63.6 - 32.25 = 31.35 \, \text{N} \]
Acceleration: \[ a = \frac{31.35}{10.0 + 16.0} = \frac{31.35}{26.0} \approx 1.206 \, \text{m/s}^2 \]
Tension in the rope: \[ T = m_1 a + f_{k1} = 10.0 \times 1.206 + 8.73 \approx 12.06 + 8.73 = 20.79 \, \text{N} \]

\[ \boxed{a \approx 1.206 \, \text{m/s}^2} \] \[ \boxed{T \approx 20.79 \, \text{N}} \]

Was this solution helpful?