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The systems shown below are in equilibrium (with m =

6.50 kg and ? = 28.0�). If the spring scales are

calibrated in newtons, what do they read? Ignore the masses of the

pulleys and strings and assume the pulleys and the incline are

frictionless.

scale in (a) ____

N scale in (b) ____

N scale in (c) ____

N scale in (d) ___N

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Added Fri, 08 Jul '16
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In equlibrium situation:

(a)

The tension in both the threads

connected to the pully is equal to mg. ( mass is m and

gravity is g )

mg = (6.5 kg) (9.81 m/s2)= 63.76N

If round off to 3 significant digits,

spring reads as 63.8 N

(b) This is similar to the previous

case

Thus the spring reads 63.76 N

If round off to 3 significant digits,

spring reads as 63.8 N

(c)

The tension ineach string s will be

mg and hence, the tension in strings connecting the

springs will be 2mg

Therefore, the spring reads 2(63.76)=

127.53 N

If round off to 3 significant digits,

spring reads as 128 N

(d)

The tension in the strings will be

equal to

mgsin? = 63.76N

(sin(28o)

= 29.93 N

If round off to 3 significant digits,

spring reads as 29.9N

Edit
Added Fri, 08 Jul '16
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