A 1019kg shark is supported by an 85kg rope attached to a 4.00-m rod that can pivot at the base.
(a) Calculate the tension in the cable between the rod and the wall, assuming that the cable is holding the system in the position shown in Figure below.
Find
(b) the horizontal force
and (c) the vertical force exerted on the base of the rod.
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To solve this problem, let's start by analyzing the forces acting on the system.
(a) Tension in the cable:
To find the tension in the cable between the rod and the wall, we need to consider the torque balance around the pivot point of the rod. The torque exerted by an object is given by the product of its weight and the perpendicular distance from the pivot point.
In this case, the weight of the shark (1019 kg) creates a torque around the pivot point, and the tension in the cable also creates a torque. Since the system is in equilibrium, both torques must be equal.
The torque created by the shark is given by:
Torque_shark = weight_shark * perpendicular distance from the pivot = m_shark * g * d
where m_shark is the mass of the shark, g is the acceleration due to gravity (approximately 9.8 m/s^2), and d is the perpendicular distance between the pivot point and the line of action of the weight of the shark.
The torque created by the tension in the cable is given by:
Torque_cable = tension_cable * perpendicular distance from the pivot = T * L
where T is the tension in the cable and L is the length of the cable from the pivot point.
Since the torques are equal in equilibrium, we can set up the equation:
m_shark * g * d = T * L
Substituting the given values:
1019 kg * 9.8 m/s^2 * d = T * 4.00 m
Solving for T, we get:
T = (1019 kg * 9.8 m/s^2 * d) / 4.00 m
Now, we need to find the perpendicular distance d. Since the rope is attached to the rod, and the rod is perpendicular to the wall, the distance d is equal to the length of the rod. Therefore, d = 4.00 m.
Substituting this value into the equation, we get:
T = (1019 kg * 9.8 m/s^2 * 4.00 m) / 4.00 m
Calculating this expression, we find the tension in the cable:
T = 10095.2 N
So, the tension in the cable between the rod and the wall is 10095.2 N.
(b) Horizontal force:
The horizontal force exerted on the base of the rod can be determined by resolving the forces in the horizontal direction. Since the system is in equilibrium, the horizontal force must balance out the horizontal component of the tension in the cable.
The horizontal component of the tension in the cable can be found using trigonometry. We can express it as:
Horizontal component of tension = T * cos(θ)
where θ is the angle between the cable and the horizontal direction.
Since the force exerted on the base of the rod balances out the horizontal component of the tension, we have:
Horizontal force = T * cos(θ)
(c) Vertical force:
The vertical force exerted on the base of the rod can be determined by resolving the forces in the vertical direction. Since the system is in equilibrium, the vertical force must balance out the weight of the shark.
The weight of the shark can be expressed as:
Weight_shark = m_shark * g
Since the force exerted on the base of the rod balances out the weight of the shark, we have:
Vertical force = Weight_shark
To summarize:
(a) The tension in the cable between the rod and the wall is 10095.2 N.
(b) The horizontal force exerted on the base of the rod is T * cos(θ).
(c) The vertical force exerted on the base of the rod is the weight of the shark, which is m_shark * g.