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Hey, i don't get either of these questions, if someone could help me I'd really appreciate it!

[this ones a pulley system problem...]
The weight of a block on a table is 422N and the weight of the block hanging is 185N. Ignoring all frictional effects and assuming the pulley to be massless, find
a) the acceleration of the two blocks
b) the tension in the cord

A 95kg person stands on a scale in an elevator
What is the apparent weight when the elevator is
a) accelerating upward with an acceleration of 1.8 m/s^2
b) moving upward at a constant speed
c) accelerating downward with an acceleration of 1.3m/s^2

thanks!

2007-10-21 01:44:07 · 3 answers · asked by Melissa S 1 in Science & Mathematics Physics

I forgot to add that the textbook has answers at the back, and for the first question with the pulley system it says the acceleration is 2.99 m/s^2 and the tension is 129N

2007-10-21 02:43:51 · update #1

3 answers

( 1 )
a) As I understand, the heavier block of 422 N on one side of the pulley is resting on the table and the lighter block of 185 N is hanging at the other end of the cord. There can be no motion and hence, acceleration is zero.

b) As the 185 N block is at rest, tension in the cord is also 185 N.

2)
a) Apparent wt. is the force of action/reaction between the person and the scale. Two forces act on the person:
( i ) his weight, mg, downwards and
( ii ) normal reaction from the sacle (apparent weight), W, upwards
As the person is accelerating upwards with the lift with acceleration a, the net upward force,
F - mg = ma
=> F = m(g + a) = 95(9.8 + 1.8) = 1102 N.

b) Constant speed means a = 0.
=> F = mg = 95 x 9.8 = 931 N.

c) For downward acceleration, a = - 1.3 m/s^2
=> F = m (g - a) = 95(9.8 - 1.3) = 807.5 N.

2007-10-21 03:29:10 · answer #1 · answered by Madhukar 7 · 0 0

In any frictionless mass-less pulley system, the tension on the 'rope' will be the same along its entire length. Therefore you must determine the unbalanced force that the two weights create. Given the unbalanced force F and knowing the total mass (of two weights) and knowing if the two masses will move at the same speed, use F = ma to calculate acceleration a.

In the second problem the elevator will add its acceleration to gravity g if going up or subtract if going down. Knowing the mass of the person, use F = ma (where g +/- a of elevator) and F (force or apparent weight of person). If the elevator is moving at constant speed it has already accelerated or decelerated and F = mg.

2007-10-21 09:32:22 · answer #2 · answered by Kes 7 · 0 0

OK. Since the two blocks are connected together, they'll accelerate at the same rate. Since the cord supplies -all of the force on the larger block, the tension of the cord is 185 N. Now, a weight of 422N has a mass of 422/9.8 = 43 kg (remember F=ma) and so a force of 185 N will cause it to accelerate at 185/43 or about 4.3 m/s². And this is also the acceleration of the other block since they're connected together.

Since in the elevator, the total acceleration is 9.8 + 1.8 = 11.6 m/s² the apparent weight of the person will be 11.6/9.8 = 1.18 times their normal weight (or about 112 kg). If the elevator is moving at constant speed, their weight will be normal. And if accelerating downwards at 1.3 m/s² the total acceleration on them will be
9.8 - 1.3 = 8.5 m/s² and their apparent weight is 8.5/9.8 = .87 times their 'normal' weight (or about 82.6 kg)

Doug

2007-10-21 09:30:06 · answer #3 · answered by doug_donaghue 7 · 1 0

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