Answer:
A 200mm lever and a 240 mm diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 860N vertical load is applied at A when the lever is horizontal, determine (a)the tension in the cord, (b)the reactions at C and D. assume that the bearing at D does not exert axial thrust.
see explanation
Explanation:
Part a
The tension in the cord
860(200) - T(120) = 0
T(120) = 172000
T =1433.33N
The tension in the cord (T) = 1433.33N
Part b
Apply the moment law of equilibrium at point D about y-axis.
sustitute 1433.33N for T
The reaction at C along x-axis 477.78N
Apply the moment law of equilibrium at point D about x-axis
The reaction at C along y-axis is 143.33N
Apply the force law of equilibrium along z direction.
The reaction at C along z-axis is 0N
Apply the force law of equilibrium along x direction.
substitute 477.78N for and 143.33N for
The reaction at D along x-axis is -1911.11 N
Apply the force law of equilibrium along y direction.
substitute 1433.33 for
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The gravitational field strength is approximately equal to 10 N.
<u>Explanation:</u>
Gravitational field strength is the measure of gravitational force acting on any object placed on the surface of the planet. Generally, the mass of the object is considered as 1 kg.
So the gravitational field strength will be equal to the gravitational force acting on the object.
The formula for gravitational field strength is
Here g is the gravitational field strength, m is the mass of the object placed on the surface and F is the gravitational force acting on the object.
Since, the mass of any object placed on the surface of earth will be negligible compared to the mass of Earth, so the mass of the object is considered as 1 kg.
Then the g = F
And
Here G is the gravitational constant, M is the mass of Earth and m is the mass of the object placed on the surface, while r is the radius of the Earth.
So, the gravitational field strength is approximately equal to 10 N.
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