Engineering Mechanics - KOP: Force and Acceleration

Exercise : KOP: Force and Acceleration - General Questions
  • KOP: Force and Acceleration - General Questions
1.

Rod OA rotates counterclockwise with a constant angular rate of = 5 rad/s. The double collar B is pin-connected together such that one collar slides over the rotating rod and the other slides over the horizontal curved rod, of which the shape is a limacon described by the equation r - 1.5(2 - cos ) ft. If both collars weigh 0.75 lb, determine the normal force which the curved path exerts on one of the collars, and the force that OA exerts on the other collar at the instant = 90°.

FOA = 0.873 lb, Fcurve = 1.953 lb
FOA = 0, Fcurve = 1.953 lb
FOA = 1.747 lb, Fcurve = 0
FOA = 2.87 lb, Fcurve = 6.41 lb
Answer: Option
Explanation:
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2.

The 30-lb crate is being hoisted upward with a constant acceleration of 6 ft/s2. If the uniform beam AB has a weight of 200 lb, determine the components of reaction at A. Neglect the size and mass of the pulley at B.

Ax = -48.3 lb, Ay = 248.3 lb, MA = 258 lb-ft CCW
Ax = -48.3 lb, Ay = 248.3 lb, MA = 741 lb-ft CCW
Ax = -35.6 lb, Ay = 236 lb, MA = 678 lb-ft CCW
Ax = -30.0 lb, Ay = 230 lb, MA = 650 lb-ft CCW
Answer: Option
Explanation:
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3.

The 300-kg bar B, originally at rest, is being towed over a series of small rollers. Computer the force in the cable when t = 5s, if the motor M is drawing in the cable for a short time at a rate of v = (0.4t2) m/s, where t is in seconds (0 t 6 s). How far does the bar move in 5 s? Neglect the mass of the cable, pulley P, and the rollers.

T = 5.00 kN, s = 0.300 m
T = 1.200 kN, s = 1.25 m
T = 5.00 kN, s = 4.00 m
T = 1.200 kN, s = 16.67 m
Answer: Option
Explanation:
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4.

A 1.5-lb brick is released from rest A and slides down the inclined roof. If the coefficient of friction between the roof and the brick is = 0.3, determine the speed at which the brick strikes the gutter G.

v = 3.00 ft/s
v = 2.68 ft/s
v = 5.61 ft/s
v = 15.23 ft/s
Answer: Option
Explanation:
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5.

A ball having a mass of 2 kg slides without friction within a vertical circular slot. If it is released from rest when = 10°, determine the force it exerts on the slot when it arrives at points A and B.

NA = 38.6 N, NB = 96.6 N
NA = 30.9 N, NB = 61.8 N
NA = 30.9 N, NB = 81.2 N
NA = 38.6 N, NB = 77.3 N
Answer: Option
Explanation:
No answer description is available. Let's discuss.