Question

ws#5 - two objects & tension
directions: draw a fbd and label all forces involved and your fnet equation. do on separate paper!
remember:

• acceleration is determined with the whole system (all masses)
• tension is determined with just one mass (use only 1 part of diagram)
• tension should be equal for the system (ie. t1 = t2 so they cancel each other out)
• redraw your fbd in a horizontal plane to make it easier.

i. determine the acceleration and tension in each system below.

Explanation:

Step1: Consider the whole - system for acceleration

For a system of connected masses, we use Newton's second law $F_{net}=ma$. The net - force acting on the whole system is determined by the unbalanced external forces. The total mass of the system is the sum of all masses involved.

Step2: Consider a single mass for tension

To find the tension in the string, we isolate one of the masses and analyze the forces acting on it. For a mass $m$ in the system, the net - force acting on it is related to its acceleration $a$ by $F_{net}=ma$. The tension is one of the forces in the free - body diagram of the single mass.

Let's take part (b) as an example:
1. **Acceleration of the system**:
- The net - force acting on the whole system is due to the weight of the hanging mass. The mass of the hanging mass $m_1 = 12\ kg$ and the mass on the table $m_2=10\ kg$. The force due to gravity on the hanging mass is $F = m_1g$, where $g = 9.8\ m/s^2$.
- According to Newton's second law for the whole system $F=(m_1 + m_2)a$. So, $m_1g=(m_1 + m_2)a$.
- Substitute $m_1 = 12\ kg$, $m_2 = 10\ kg$ and $g = 9.8\ m/s^2$ into the equation: $12\times9.8=(12 + 10)a$.
- Then $a=\frac{12\times9.8}{12 + 10}=\frac{117.6}{22}\approx5.35\ m/s^2$.
2. **Tension in the string**:
- Consider the mass on the table $m_2 = 10\ kg$. The only horizontal force acting on it is the tension $T$. According to Newton's second law $T=m_2a$.
- Substitute $m_2 = 10\ kg$ and $a\approx5.35\ m/s^2$ into the equation, we get $T = 10\times5.35 = 53.5\ N$.

We can follow the same procedure for parts (a), (c) and (d).

For part (a):
Let the two hanging masses be $m_1 = 12\ kg$ and $m_2 = 10\ kg$.
1. **Acceleration**:
- The net - force on the system is $F=(m_1 - m_2)g$ (assuming the heavier mass is pulling the lighter one). The total mass of the system is $m_{total}=m_1 + m_2$.
- By Newton's second law $(m_1 - m_2)g=(m_1 + m_2)a$. Substitute $m_1 = 12\ kg$, $m_2 = 10\ kg$ and $g = 9.8\ m/s^2$.
- $(12 - 10)\times9.8=(12 + 10)a$. So, $a=\frac{2\times9.8}{22}\approx0.89\ m/s^2$.
2. **Tension**:
- Consider the $10\ kg$ mass. The force equation is $T - m_2g=m_2a$. Rearranging for $T$, we get $T=m_2(g + a)$. Substitute $m_2 = 10\ kg$, $g = 9.8\ m/s^2$ and $a\approx0.89\ m/s^2$. Then $T=10\times(9.8 + 0.89)=10\times10.69 = 106.9\ N$.

For part (c):
The mass on the table $m_1 = 40\ kg$ and the hanging mass $m_2 = 25\ kg$, and the coefficient of friction $\mu=0.40$.
1. **Acceleration**:
- The force of friction on the mass on the table is $F_f=\mu N=\mu m_1g$, where $N = m_1g$ (since the mass is on a horizontal surface). The net - force on the system is $F = m_2g-\mu m_1g$. The total mass of the system is $m_{total}=m_1 + m_2$.
- By Newton's second law $m_2g-\mu m_1g=(m_1 + m_2)a$.
- Substitute $m_1 = 40\ kg$, $m_2 = 25\ kg$, $\mu = 0.40$ and $g = 9.8\ m/s^2$.
- $25\times9.8-0.4\times40\times9.8=(40 + 25)a$.
- $245-156.8 = 65a$.
- $88.2 = 65a$, so $a=\frac{88.2}{65}\approx1.36\ m/s^2$.
2. **Tension**:
- Consider the hanging mass $m_2$. The force equation is $m_2g - T=m_2a$. Rearranging for $T$, we get $T=m_2(g - a)$. Substitute $m_2 = 25\ kg$, $g = 9.8\ m/s^2$ and $a\approx1.36\ m/s^2$. Then $T=25\times(9.8 - 1.36)=25\times8.44 = 211\ N$.

For part (d):
Let the two hanging masses be $m_1 = 15\ kg$ and $m_2 = 10\ kg$ and the mass on the table $m_3 = 5.0\ kg$.
1. **Acceleration**:
- The net - force on the system is $(m_1 - m_2)g$. The total mass of the system is $m_{total}=m_1 + m_2+m_3$.
- By Newton's second law $(m_1 - m_2)g=(m_1 + m_2+m_3)a$. Substitute $m_1 = 15\ kg$, $m_2 = 10\ kg$, $m_3 = 5.0\ kg$ and $g = 9.8\ m/s^2$.
- $(15 - 10)\times9.8=(15 + 10+5)a$.
- $49 = 30a$, so $a=\frac{49}{30}\approx1.63\ m/s^2$.
2. **Tension**:
- Consider the $10\ kg$ mass. The force equation is $T - m_2g=m_2a$. Rearranging for $T$, we get $T=m_2(g + a)$. Substitute $m_2 = 10\ kg$, $g = 9.8\ m/s^2$ and $a\approx1.63\ m/s^2$. Then $T=10\times(9.8 + 1.63)=10\times11.43 = 114.3\ N$.

Answer:

a. Acceleration: $a\approx0.89\ m/s^2$, Tension: $T\approx106.9\ N$
b. Acceleration: $a\approx5.35\ m/s^2$, Tension: $T = 53.5\ N$
c. Acceleration: $a\approx1.36\ m/s^2$, Tension: $T\approx211\ N$
d. Acceleration: $a\approx1.63\ m/s^2$, Tension: $T\approx114.3\ N$