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Juli2301 [7.4K]
3 years ago
14

Suppose you have a 115 kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood sur

faces. (Enter the magnitudes.) (a) What maximum force (in N) can you exert horizontally on the crate without moving it
Physics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

maximum force (in N) can you exert horizontally on the crate without moving it = 564.075 N

Explanation:

Given data:

mass m= 115 kg

coefficient of friction μ =0.5

according to newton law of motion the net force acted on body is given by

F_{net}=ma=F_{max}- F_{f}

here the body is in rest ,

F_{net}=0\\\Rightarrow F_{max}=F_{f}\\\text{frriction force}  F_{f} = \mu\times N\\

N= it is the normal reaction force

N= mg = 115×9.81

therefore,

F_{f} = 0.5\times115\times9.81 = 564.075 N = F_{max}

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density of water, \rho _w = 1000 kg/m³

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Based on the density of the wood, it will position across the two liquids.

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Let the wood be in equilibrium position;

F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood}  -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood}  -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

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