Question Number 74522 by necxxx last updated on 25/Nov/19
$${A}\:{rope}\:{inclined}\:{at}\:{angle}\:\mathrm{37}°\:{to}\:{the}\: \\ $$$${horizontal}\:{is}\:{used}\:{to}\:{drag}\:{a}\:\mathrm{50}{kg}\:{block} \\ $$$${along}\:{a}\:{level}\:{floor}\:{with}\:{an}\:{acceleration} \\ $$$${of}\:\mathrm{1}{m}/{s}^{\mathrm{2}} \:.{The}\:{coefficient}\:{of}\:{friction} \\ $$$${between}\:{the}\:{block}\:{and}\:{the}\:{floor}\:{is}\:\mathrm{0}.\mathrm{2}. \\ $$$${What}\:{is}\:{the}\:{tension}\:{in}\:{the}\:{rope}? \\ $$
Commented by Rio Michael last updated on 25/Nov/19
Commented by Rio Michael last updated on 25/Nov/19
$${from}\:{the}\:{above}\:{diagram}, \\ $$$$\:\:{Tcos}\:\mathrm{37}°\:−\:{F}_{{f}} \:=\:{ma} \\ $$$$\:\:{Tcos}\mathrm{37}°\:−\:{F}_{{N}} \mu\:=\:{ma} \\ $$$${but}\:\:{F}_{{N}} \:=\:{W}\:=\:{mg} \\ $$$$\therefore\:\:{Tcos}\mathrm{37}\:−\:{mg}\mu\:=\:{ma} \\ $$$$\Rightarrow\:{T}\:=\:\frac{{m}\left({a}\:+\:{g}\mu\right)}{{cos}\mathrm{37}°}\: \\ $$$$\:\:\:\:\:{T}\:=\:\frac{\mathrm{50}\left(\:\mathrm{1}\:+\:\mathrm{10}\left(\mathrm{0}.\mathrm{2}\right)\right)}{{cos}\mathrm{37}}\:=\:\mathrm{187}.\mathrm{8}{N} \\ $$
Commented by necxxx last updated on 25/Nov/19
$${Thank}\:{you}\:{sir}\:{but}\:{the}\:{book}\:{says}\:{that}\:{the} \\ $$$${answer}\:{is}\:\mathrm{161}{N} \\ $$