A 207-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 31o with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.87, and the log has an acceleration of 0.87 m/s^2 . Find the tension in the rope.
2007-06-20 16:53:39 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
There are 3 forces acting on the log - the force of gravitation, the force of friction and the tension of the rope.
Gravitational force = mg sin θ = 207 * 9.8 * sin 31 = 1044.8 N
Force of friction = μ mg cos θ = 0.87 * 207 * 9.8 * cos 31 = 1515.80
The tension in the string acts opposite to the combined might of these 2 forces and there is also an acceleration of 0.87 m/s
Using Newton's second law
T - (1515.80 + 1044.8) = ma
T - (2560.6) = 207 * 0.87
T = 2560.6 + 180.09
T = 2740.69 N
2007-06-20 17:31:39 · answer #1 · answered by Ajinkya N 5 · 0⤊ 0⤋
T = ma + μmgcosθ + mgsinθ
T = m(a + μgcosθ + gsinθ)
T = 207(0.87 + 9.80665(0.87cos31° + sin31°))
T = 207(0.87 + 9.80665(0.87(0.8571673) + 0.5150381))
T = 207(0.87 + 9.80665(0.7457356 + 0.5150381))
T = 207(0.87 + 9.80665(1.260774))
T = 207(0.87 + 12.36397)
T = 207(13.23397)
T = 2739.431 N
2007-06-21 01:01:17 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
T = Friction + 207(9.8)sin31
I forgot the friction formula.
2007-06-21 00:29:34 · answer #3 · answered by Math Master 2 · 0⤊ 0⤋