A 4.0 m long steel beam with a cross-sectional area of 1.0 x102m2 and a Young's modulus
of 2.0 x10' N/m 2 is wedged horizontally between two vertical walls. In order to wedge the
beam, it is compressed by 0.020 mm. If the coffecient of static friction between the beam
and the walls is 0.70, the maximum mass (including its own) it can bear without slipping
is:
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Ответ:
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.
Explanation:
Measure unstretched length of spring, L. E.g. L = 0.60m.
Set mass to a convenient value (e.g. m = 0.5kg).
Hang mass.
Measure new spring length, L'. E.g. L' = 0.70m.
Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m
Use mg = ke (in equilibrium weight = tension)
k = mg/e
Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).
k = 0.5*10/0.10 = 50 N/m
Repeat for a few different masses. (L always stays the same.)
Take the average of your k values.