If $$ \overrightarrow{\mathrm{F}}_{1}, \overrightarrow{\mathrm{~F}}_{2} $$ are two unlike parallel forces act at the two points $$ \mathrm{A}, \mathrm{B} $$ respectively such that $$ \mathrm{AB}=12.5 \mathrm{~cm},, \mathrm{~F}_{1}=80 $$ dyne, $$ \mathrm{F}_{2}=30 $$ dyne, find the resultant of the two forces. ABCD is a rectangle in which $$ \mathrm{AB}=30 \mathrm{~cm}, \mathrm{BC}=40 \mathrm{~cm} $$. forces of magnitudes 15,30 , 15,30 gm.wt. act along $$ \overrightarrow{\mathrm{BA}}, \overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{DC}}, \overrightarrow{\mathrm{DA}} $$ respectively. Prove that this system of forces is equivalent to a couple and find the magnitude of its moment , then find two forces acting at A and C perpendicular to $$ \overrightarrow{\mathrm{AC}} $$ such that the system is in equilibrium.

Question

UpStudy AI · Accepted Answer

To solve the problem, we will break it down into two parts:

1. Finding the resultant of the two unlike parallel forces $$ \overrightarrow{F}_{1} $$ and $$ \overrightarrow{F}_{2} $$.
2. Proving that the system of forces acting on the rectangle ABCD is equivalent to a couple and finding the moment of the couple, then determining the forces acting at points A and C.

### Part 1: Resultant of the Two Forces

Given:
- Distance $$ AB = 12.5 $$ cm
- Force $$ F_{1} = 80 $$ dyne (acting at point A)
- Force $$ F_{2} = 30 $$ dyne (acting at point B)

The resultant $$ R $$ of two unlike parallel forces can be calculated using the formula:

$$
R = F_{1} - F_{2}
$$

Substituting the values:

$$
R = 80 \, \text{dyne} - 30 \, \text{dyne} = 50 \, \text{dyne}
$$

### Part 2: Proving the System of Forces is Equivalent to a Couple

For the rectangle ABCD, the forces acting are:
- $$ F_{BA} = 15 $$ gm.wt. (acting along $$ \overrightarrow{BA} $$)
- $$ F_{BC} = 30 $$ gm.wt. (acting along $$ \overrightarrow{BC} $$)
- $$ F_{DC} = 15 $$ gm.wt. (acting along $$ \overrightarrow{DC} $$)
- $$ F_{DA} = 30 $$ gm.wt. (acting along $$ \overrightarrow{DA} $$)

To prove that this system is equivalent to a couple, we need to check if the sum of the vertical forces and the sum of the horizontal forces are equal and opposite.

1. **Vertical Forces:**
   - Upward forces: $$ F_{BC} + F_{DA} = 30 + 30 = 60 $$ gm.wt.
   - Downward forces: $$ F_{BA} + F_{DC} = 15 + 15 = 30 $$ gm.wt.

The net vertical force is $$ 60 - 30 = 30 $$ gm.wt. (upward).

2. **Horizontal Forces:**
   - Leftward forces: $$ F_{BA} = 15 $$ gm.wt.
   - Rightward forces: $$ F_{DC} = 15 $$ gm.wt.

The net horizontal force is $$ 15 - 15 = 0 $$.

Since the vertical forces do not balance, we need to calculate the moment about a point (let's say point A) to find the moment of the couple.

### Moment of the Couple

The moment $$ M $$ about point A due to the forces can be calculated as:

$$
M = F \times d
$$

Where $$ d $$ is the perpendicular distance from the line of action of the force to the point A.

- For $$ F_{BC} $$ (30 gm.wt) acting at B (distance $$ AB = 30 $$ cm):
  $$
  M_{BC} = 30 \times 30 = 900 \, \text{gm.cm}
  $$

- For $$ F_{DA} $$ (30 gm.wt) acting at D (distance $$ AD = 40 $$ cm):
  $$
  M_{DA} = 30 \times 40 = 1200 \, \text{gm.cm}
  $$

The total moment about point A is:

$$
M = M_{BC} + M_{DA} = 900 + 1200 = 2100 \, \text{gm.cm}
$$

### Finding Forces at A and C

To maintain equilibrium, we need to find two forces $$ F_{A} $$ and $$ F_{C} $$ acting at points A and C, respectively, perpendicular to $$ \overrightarrow{AC} $$.

Let $$ F_{A} $$ and $$ F_{C} $$ be the forces acting at A and C. The system will be in equilibrium if:

1. The sum of vertical forces equals zero.
2. The sum of moments about any point equals zero.

Assuming $$ F_{A} $$ acts upward and $$ F_{C} $$ acts downward, we can set up the equations:

1. $$ F_{A} - F_{C} + 30 = 0 $$ (for vertical forces)
2. The moment about point A due to $$ F_{C} $$ must equal the moment due to the other forces.

Let $$ d_{AC} $$ be the distance between A and C, which is the diagonal of the rectangle:

$$
d_{AC} = \sqrt{(30^2 + 40^2)} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{cm}
$$

The moment due to $$ F_{C} $$ about A is:

$$
M_{C} = F_{C} \times 50
$$

Setting the moments equal for equilibrium:

$$
F_{C} \times 50 = 2100
$$

Solving for $$ F_{C} $$:

$$
F_{C} = \frac{2100}{50} = 42 \, \text{gm.wt}
$$

Now substituting back to find $$ F_{A} $$:

$$
F_{A} - 42 + 30 = 0 \implies F_{A} = 12 \, \text{gm.wt}
$$

### Summary of Results

1. The resultant of the two forces $$ R = 50 $$ dyne.
2. The moment of the couple $$ M = 2100 \, \text{gm.cm} $$.
3. The forces acting at A and C are $$ F_{A} = 12 \, \text{gm.wt} $$ and $$ F_{C} = 42 \, \text{gm.wt} $$, respectively.
The resultant force is 50 dyne. The system of forces forms a couple with a moment of 2100 gm.cm. To maintain equilibrium, two forces of 12 gm.wt and 42 gm.wt act at points A and C, respectively.

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Physics Questions