What is the minimum force F must be applied to lift the body mass m? The height of the inclined plane is equal to H and the length of the inclined plane is equal to L. Consider the blocks to be weightless and neglect friction.

Solution. Posted on Saturday, December 27, 2014.

If a movable pulley moved a distance l than the force F performed work equal to F*l. The body on the inclined plane at the same time moved a distance 2*l and rose to a height of h. The amount of gravitational potential energy (m*g*h) possessed by an elevated the body is equal to the work done. Thereby, m*g*h=F*l. The h could be found from the similarity of triangles.

h = 2*l*H/L

Now, when looking at the previous equation, it can be written as m*g*2*l*H/L=F*l

Finally,

F=2*m*g*H/L

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First find the angle, theta, between the plane

and floor.

theta = arcsin(H/L)

The force, f, against the rope parallel

to the incline is

f = m*g*cos(theta)

which is the force needed to lift it up

the incline. The pulley at the end of the

incline doesn’t multiply the force f, it

just transmits it to one side of the other

pulley.

The minimum force, F, to lift the mass, m,

is twice the two forces pulling up of the

pulley pulling down. The two forces lifting

up are equal to f.

F = 2*m*g*cos(theta).

It’s hard to answer this problem with mostly

words, but fun.

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Thank you for your solution but

the Answer is F = 2mgH/L

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I see, looking at it again. It’s the sin, not the cosine,

and sin(arcsin(H/L)) gives H/L. Thanks.

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