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A-rope-inclined-at-angle-37-to-the-horizontal-is-used-to-drag-a-50kg-block-along-a-level-floor-with-an-acceleration-of-1m-s-2-The-coefficient-of-friction-between-the-block-and-the-floor-is-0-2-Wh




Question Number 74522 by necxxx last updated on 25/Nov/19
A rope inclined at angle 37° to the   horizontal is used to drag a 50kg block  along a level floor with an acceleration  of 1m/s^2  .The coefficient of friction  between the block and the floor is 0.2.  What is the tension in the rope?
$${A}\:{rope}\:{inclined}\:{at}\:{angle}\:\mathrm{37}°\:{to}\:{the}\: \\ $$$${horizontal}\:{is}\:{used}\:{to}\:{drag}\:{a}\:\mathrm{50}{kg}\:{block} \\ $$$${along}\:{a}\:{level}\:{floor}\:{with}\:{an}\:{acceleration} \\ $$$${of}\:\mathrm{1}{m}/{s}^{\mathrm{2}} \:.{The}\:{coefficient}\:{of}\:{friction} \\ $$$${between}\:{the}\:{block}\:{and}\:{the}\:{floor}\:{is}\:\mathrm{0}.\mathrm{2}. \\ $$$${What}\:{is}\:{the}\:{tension}\:{in}\:{the}\:{rope}? \\ $$
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 37° − F_f  = ma    Tcos37° − F_N μ = ma  but  F_N  = W = mg  ∴  Tcos37 − mgμ = ma  ⇒ T = ((m(a + gμ))/(cos37°))        T = ((50( 1 + 10(0.2)))/(cos37)) = 187.8N
$${from}\:{the}\:{above}\:{diagram}, \\ $$$$\:\:{Tcos}\:\mathrm{37}°\:−\:{F}_{{f}} \:=\:{ma} \\ $$$$\:\:{Tcos}\mathrm{37}°\:−\:{F}_{{N}} \mu\:=\:{ma} \\ $$$${but}\:\:{F}_{{N}} \:=\:{W}\:=\:{mg} \\ $$$$\therefore\:\:{Tcos}\mathrm{37}\:−\:{mg}\mu\:=\:{ma} \\ $$$$\Rightarrow\:{T}\:=\:\frac{{m}\left({a}\:+\:{g}\mu\right)}{{cos}\mathrm{37}°}\: \\ $$$$\:\:\:\:\:{T}\:=\:\frac{\mathrm{50}\left(\:\mathrm{1}\:+\:\mathrm{10}\left(\mathrm{0}.\mathrm{2}\right)\right)}{{cos}\mathrm{37}}\:=\:\mathrm{187}.\mathrm{8}{N} \\ $$
Commented by necxxx last updated on 25/Nov/19
Thank you sir but the book says that the  answer is 161N
$${Thank}\:{you}\:{sir}\:{but}\:{the}\:{book}\:{says}\:{that}\:{the} \\ $$$${answer}\:{is}\:\mathrm{161}{N} \\ $$

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