You are an engineer working for a manufacturing company. You are designing a mec

Question

You are an engineer working for a manufacturing company. You are designing a mechanism that uses a cable to drag heavy metal blocks a distance of 8.00 m along a ramp that is sloped at 40.0 above the horizontal. The coefficient of kinetic friction between these blocks and the incline is k = 0.350. Each block has a mass of 2090 kg . The block will be placed on the bottom of the ramp, the cable will be attached, and the block will then be given just enough of a momentary push to overcome static friction. The block is then to accelerate at a constant rate to move the 8.00 m in 4.20 s. The cable is made of wire rope and is parallel to the ramp surface. The table gives the breaking strength of the cable as a function of its diameter; the safe load tension, which is 20% of the breaking strength; and the mass per meter of the cable:

D.) The coefficient of static friction between the crate and the ramp is s = 0.620, which is nearly twice the value of the coefficient of kinetic friction. If the machinery jams and the block stops in the middle of the ramp, what is the tension in the cable? Assume that the block tends to slip down the slope and the maximum static friction force acts on the block.

Explanation / Answer

Tension in the Cable = Weight of the Block acting downward along the incline - Friction Force (As it will oppose the motion.)
Tension in the Cable = M*g*sin(40) - s* M*g*cos(40)
Substituing Values

Tension in the Cable = 2090 * 9.8 * sin(40) - 0.620 * 2090 * 9.8 * cos(40)
Tension in the Cable = 3437.7 N

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