A.
B. Calculate the normal reaction:
"f=\\mu N,\\\\\\space\\\\\nN=\\frac{f}{\\mu}."On the other hand, assuming zero acceleration under 100N force and normal coordinate system (y up and x to the right), according to Newton's second law, we have
"y: N-mg+F\\text{ sin}20\u00b0=0,\n\\\\\nN=mg-F\\text{ sin}20\u00b0=750\\text{ N}.\\\\\nx: f-F\\text{cos}20\u00b0=0,\\\\\nf=F\\text{cos}20\u00b0=94\\text{ N}." C. Write the same condition for equilibrium along x-axis:
"x: f-F\\text{cos}20\u00b0=0,\\\\\nF=f\/\\text{cos}20\u00b0=130\\text{ N}." D. The acceleration is
"f-F\\text{ cos}20\u00b0=-ma,\\\\\\space\\\\\na=\\frac{F\\text{ cos}20\u00b0-f}{m}=0.14\\text{ m\/s}^2."
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