Author Topic: Dynamics (Read 14554 times) Tweet Share

d0minicz · « **Reply #75 on:** September 11, 2009, 10:50:33 pm »

A block of mass 10kg rests on the floor of a lift which is accelerating upwards at 4 metres per second squared. What is the magnitude of the reaction force of the floor of the lift on the block (assuming acceleration due to gravity has magnitude 9.8 m/s^2) ?

how do you do these lift ones lol

thanks !

danieltennis · « **Reply #76 on:** September 11, 2009, 10:56:43 pm »

ma = N - mg (take up as positive)
10*4 = N - 10*9.8
N = 138 N

d0minicz · « **Reply #77 on:** September 13, 2009, 01:57:40 am »

A mass of 500 grams hangs from a point M by a rope 8cm long. It is pulled aside by a horizontal force of P kg wt until it is 4cm from the vertical line through M. Find the value of P.

thanks

/0 · « **Reply #78 on:** September 13, 2009, 03:00:16 am »

Let the tension in the rope be T
Find the angle $\theta$ the rope makes with the horizontal using trigonometry.

Then:
$T\cos{\theta} = P$
$T\sin{\theta} = mg$

d0minicz · « **Reply #79 on:** September 13, 2009, 09:30:38 pm »

ahh can someone show me working out for it please

kamil9876 · « **Reply #80 on:** September 13, 2009, 09:48:53 pm »

GOing from where /0 left of.
$<br />cos\theta=\frac{4}{8}=\frac{1}{2} \implies \theta=60^o<br />$

$\therefore sin\theta=\frac{\sqrt{3}}{2}$

sub into /0's second equation:

$T\frac{\sqrt{3}}{2}=0.5g$
$<br />T=\frac{g\sqrt{3}}{3}$

Now sub into first equation to find P.

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