A uniform rod pivoted at one end (point O) is free to swing in a vertical plane in a grav- itational field, held in equilibrium by a force F at its other end. The rod of length 7 m and weight 4.7 N makes an angle 39◦ with the hor- izontal and the force an angle of 54◦ with the horizontal
What is the magnitude of the force F ?
To find the magnitude of the force F, we can break down the problem into its horizontal and vertical components.
1. Horizontal Component:
Since the rod is in equilibrium, the sum of the horizontal components of the forces must be zero. The only force with a horizontal component is force F.
Therefore, we can calculate the horizontal component of force F using the formula:
F_horizontal = F * cos(angle with the horizontal)
Given that the angle of force F with the horizontal is 54 degrees, we can substitute it into the formula:
F_horizontal = F * cos(54 degrees)
2. Vertical Component:
Similarly, the sum of the vertical components of the forces must also be zero. The only force with a vertical component is the weight of the rod.
The weight of the rod can be calculated using the formula:
Weight = mass * gravitational acceleration
However, we are given the weight directly as 4.7 N. So we can use this value as the vertical component of the force.
3. Equilibrium Conditions:
Since the rod is in equilibrium, the horizontal and vertical components of forces must balance each other.
So we can write two equations based on these conditions:
F_horizontal = 0 (Horizontal force equilibrium)
F_vertical + Weight = 0 (Vertical force equilibrium)
4. Solving for F:
Using the equations, we can substitute the corresponding values:
F * cos(54 degrees) = 0 (Horizontal force equation)
F + 4.7 N = 0 (Vertical force equation)
To solve this system of equations, we can isolate F in the second equation:
F = -4.7 N
Since force can not be negative, this means there is an error in the problem statement or calculations.
Please double-check the given values and confirm if the values are correct.