Hopeful for greater opportunities, Juan applies for an international scholarship. Fortunately, he got accepted and is now at the airport to travel to New Zealand. Waiting on queue at the entrance, Juan is holding his 30kg luggage up on a ramp that is inclined at an angle of 10° from the horizontal. He is holding on to his luggage at an angle of 65° from the ramp. Compute for the tension the luggage exerts on his hand and the normal force.

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter4: The Laws Of Motion
Section: Chapter Questions
Problem 23P: A bag of cement weighing 325 N hangs in equilibrium from three wires as suggested in Figure P4.23....
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Hopeful for greater opportunities, Juan applies for an international scholarship. Fortunately, he got
accepted and is now at the airport to travel to New Zealand. Waiting on queue at the entrance, Juan
is holding his 30kg luggage up on a ramp that is inclined at an angle of 10° from the horizontal.
He is holding on to his luggage at an angle of 65° from the ramp. Compute for the tension the
luggage exerts on his hand and the normal force.
Transcribed Image Text:Hopeful for greater opportunities, Juan applies for an international scholarship. Fortunately, he got accepted and is now at the airport to travel to New Zealand. Waiting on queue at the entrance, Juan is holding his 30kg luggage up on a ramp that is inclined at an angle of 10° from the horizontal. He is holding on to his luggage at an angle of 65° from the ramp. Compute for the tension the luggage exerts on his hand and the normal force.
Solving for W
W = 30kg x 9.81m/s?
W = 294.3N
Solving for T & N
E Fx = 0
Tcos(75°) + Ncos(100°) + Wcos(270°) = 0
Tcos(75°) + Ncos(100°) = –Wcos(270°)
%3D
E Fy = 0
Tsin(75°) + Nsin(100°) + Wsin(270°) = 0
Tsin(75°) + Nsin(100°) = –Wsin(270°)
T = 120.923924N
N = 180.234628N
Transcribed Image Text:Solving for W W = 30kg x 9.81m/s? W = 294.3N Solving for T & N E Fx = 0 Tcos(75°) + Ncos(100°) + Wcos(270°) = 0 Tcos(75°) + Ncos(100°) = –Wcos(270°) %3D E Fy = 0 Tsin(75°) + Nsin(100°) + Wsin(270°) = 0 Tsin(75°) + Nsin(100°) = –Wsin(270°) T = 120.923924N N = 180.234628N
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