A massless rod of length L = 3.6 m stands up straight, fixed to the ground by a bolt. A horizontal force of 5.5 N is applied at a vertical distance of L/2 to the right. To counter this force and keep the rod stationary, a wire is fixed at the top of the rod and attached to the ground some distance away to the left, making an angle of 45 degrees to the horizontal.
(a)
What is the tension in the wire, in N?
(b)
Calculate the horizontal component of the force of the bolt on the rod, in N, taking right as positive.
(c)
Calculate the vertical component of the force of the bolt on the rod, in N, taking up as positive.
To find the tension in the wire, we can use the principle of torque equilibrium. The sum of the torques about any point on the rod must equal zero for it to remain stationary.
(a) To calculate the tension in the wire, we need to consider the torques acting on the rod. The force of 5.5 N applied at a vertical distance L/2 to the right will create a torque in the counterclockwise direction. The tension in the wire will create a torque in the clockwise direction to balance the torque from the applied force.
Let's choose the point where the rod is fixed to the ground as our reference point. The torque due to the applied force can be calculated as follows:
Torque = (Force) x (Distance)
Torque = 5.5 N x (L/2) = 5.5 N x (3.6 m/2) = 9.9 N·m
For the rod to remain stationary, the torque created by the tension in the wire must be equal and opposite to the torque created by the applied force. Since the distance between the point of attachment of the wire to the rod and the point where the rod is fixed to the ground is not given, we need to find it.
Given that the angle between the wire and the horizontal is 45 degrees, we can use trigonometry to find the horizontal distance between the point of attachment of the wire and the point where the rod is fixed.
Let 'x' be the horizontal distance between the points.
Using the sine function, we have:
sin(45°) = x / L
x = L x sin(45°)
x = 3.6 m x sin(45°)
x = 3.6 m x √(2)/2
x = 2.548 m (approximated to three decimal places)
Now that we have the distance 'x', we can calculate the torque due to the tension in the wire:
Torque = (Force) x (Distance)
Torque = Tension x x
Since the torques must be equal and opposite, we can equate the equations for torque and solve for the tension:
9.9 N·m = Tension x 2.548 m
Tension = 9.9 N·m / 2.548 m ≈ 3.881 N
Therefore, the tension in the wire is approximately 3.881 N.
(b) To calculate the horizontal component of the force of the bolt on the rod, we need to consider the horizontal equilibrium of forces acting on the rod.
The applied force of 5.5 N to the right creates a horizontal force component that must be balanced by the horizontal force component from the bolt. Since the rod is stationary, the net horizontal force must be zero.
Horizontal Force of the Bolt = -5.5 N (opposite in direction)
Therefore, the horizontal component of the force of the bolt on the rod is -5.5 N.
(c) To calculate the vertical component of the force of the bolt on the rod, we need to consider the vertical equilibrium of forces acting on the rod.
Since the rod is stationary, the net vertical force must be zero. The vertical component of the force of the bolt on the rod will balance the vertical component of the applied force and the weight of the rod.
The weight of the rod is equal to its mass multiplied by the acceleration due to gravity. However, since the rod is massless, the weight is zero.
Vertical Component of Applied Force = 5.5 N x cos(45°)
Vertical Component of the Force of the Bolt = 0 N (as there is no weight)
Therefore, the vertical component of the force of the bolt on the rod is 0 N.