A heavy bumper weighing 150 N is suspended along the wall of the Welland Canal to protect the ships from scraping against the side of the wall.
A sailor uses a long pole with a hook on the end to grab the rope holding the bumper and pulls it out horizontally so that the rope makes a 40 degree angle with the wall. What is the tension in the sailor's arm and the rope?
The bumper is initially hung vertically and it has a weight of 150 N.
The sailor grabs the rope holding the bumper and pulls it horizontally so that the rope makes an angle of 40 degrees with the wall.
We have two forces acting on the rope, one is by the weight of the bumper acting vertically downwards and the other is the force applied horizontally by the sailor pulling the rope. The result of the forces is that the rope makes a 40 degree incline.
Let the tension in the rope be T. This can be divided into its horizontal and vertical components.
The vertical component is T*cos 40. This is equal to the weight of the bumper.
T*cos 40 = 150
=> T = 150 / cos 40
=> T = 195.81
The horizontal component is T*sin 40. This is equal to the force being applied by the sailor.
T* sin 40 = F
We have determined that T = 195.81
=> F = 195.81* sin 40 = 125.86
The force applied by the sailor is 125.86 N and the tension in the rope is 195.81 N.