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Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 185 mm. neglect the weight of a and
b. the coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.

Respuesta :

meerkat18

By equation of equilibrium and friction:

Fb = Kx = 15(0.175) = 2.625 kN.

The wedge is on the verge of moving right then slipping will have to occur at both contact surfaces.

Fa = usNa = 0.35Na

Fb = 0.35Nb

Nb = 2.625 = 0; Nb = 2.625 kN

Nacos10 – 0.35Na sin 10 = 2.625 = 0

Na = 2.841 kN

P – (0.35 * 2.625) – 0.35 (2.841) cos 10 – 2.841 sin 10 = 0

P = 2.39 kN