Asked by ian
A tow truck is connected to a 1500 kg car by a cable that makes a 27 degree angle to the horizontal. If the truck accelerates at 0.58 m/s^2, what is the magnitude of the cable tension? Neglect friction and the mass of the cable.
Answers
Answered by
Henry
Wc = mg = 1500kg * 9.8N/kg = 14,700 N. = Wt. of car.
Fc = 14700N @ 0 Deg. = Force of car.
Fp=14,700*sin(0) = 0 = Force parallel to road.
Fv = 14,700*cos(0) = 14,700 N. = Force
perpendicular to road.
Fap*cos27 = ma = 1500*0.58 = 870 N.
Fap = 870 / cos27 = 976.4 N.=Force applied = Tension in rope.
Fc = 14700N @ 0 Deg. = Force of car.
Fp=14,700*sin(0) = 0 = Force parallel to road.
Fv = 14,700*cos(0) = 14,700 N. = Force
perpendicular to road.
Fap*cos27 = ma = 1500*0.58 = 870 N.
Fap = 870 / cos27 = 976.4 N.=Force applied = Tension in rope.
Answered by
Henry
Use only the INFO in the last 2 lines.
The other INFO is not neded.
The other INFO is not neded.
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