a uniform (in density and shape) stick is balanced by one end on the ground and the other tied to a massless horizontal rope attached to a wall as shown in the figure. The stick's mass is 10kg and its length is 5m.
a. Calculate the tension on the rope
b. Calculate the force (direction and magnitude) exerted by the stick by the ground.
To solve this problem, we can break it down into smaller parts. First, let's tackle part a.
a. To determine the tension on the rope, we need to analyze the forces acting on the stick. Since the stick is in equilibrium (it is balanced and not moving), the forces must also be in equilibrium. The two main forces acting on the stick are the gravitational force acting downward and the tension in the rope acting upward.
To find the tension on the rope, we need to consider the torque equilibrium. Torque is the product of force and the perpendicular distance from the axis of rotation.
Since the stick is balanced on one end, the axis of rotation is at the base of the stick, where it touches the ground. Let's call this point "P." The tension in the rope exerts no torque about point P because the force is acting along the axis of rotation.
Now, consider the gravitational force acting on the stick. The force of gravity acts at the center of mass of the stick, which is the midpoint of the stick. Let's call this point "C."
The distance between point P and point C is half the length of the stick, so it is 5m / 2 = 2.5m.
To maintain equilibrium, the torque exerted by the gravitational force about point P must be balanced by the torque exerted by the tension in the rope. The torque equation is given by torque = force * perpendicular distance.
Torque by tension = 0 (because it acts along the axis of rotation)
Torque by gravity = force of gravity * perpendicular distance = m * g * 2.5m
Since the stick is uniform, its mass is evenly distributed along its length. Therefore, the gravitational force acting on the stick is equal to the weight of the stick, which is given by the formula m * g.
Substituting the values, we get:
Torque by gravity = (10kg * 9.8 m/s²) * 2.5m = 245 Nm
Since the torques are equal at equilibrium, we have:
Torque by tension = Torque by gravity
0 = 245 Nm
Therefore, the tension on the rope is 0 N.
b. Since the tension on the rope is 0 N, the force exerted by the stick on the ground must equal the gravitational force acting on the stick.
The force exerted by the stick on the ground is equal in magnitude and opposite in direction to the gravitational force. The magnitude of the gravitational force is given by m * g, where m is the mass of the stick (10kg) and g is the acceleration due to gravity (9.8 m/s²).
So, the force exerted by the stick on the ground is 10kg * 9.8 m/s² = 98 N, directed vertically downward.