Asked by Kaleigh
Two objects are placed on a frictionless incline which is inclined at 65.5 degrees. A mass of m1 = 7 kg is tied to a wall, which is at the top of the incline, by rope 1 which is parallel to the incline. A mass of m2 = 13 kg is tied to m1 by rope 2 which is also parallel to the incline. Lastly a force of F is pulling m2 down the incline (parallel again). If rope 1 breaks under 300 Newtons of force, what is the largest magnitude force F can be in Newtons?
Please help me solve this !!!!!!!!!!!
Please help me solve this !!!!!!!!!!!
Answers
Answered by
bobpursley
the masses have a weight component down the incline of mg*sinTheta.
So the force due to gravity down the incline on rope 1 (if I understand your description) is (m1+m2)g*sinTheta
now, if you add F, so that F+weight=300N
F= 300N-(m1+m2)gSinTheta
So the force due to gravity down the incline on rope 1 (if I understand your description) is (m1+m2)g*sinTheta
now, if you add F, so that F+weight=300N
F= 300N-(m1+m2)gSinTheta
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