Magnitude of the tension force

A boy is pulling a sled loaded with mass m = 20 kg m = 20\text{ kg} on some horizontal, snowy pavement. The velocity of the sled is constant. The coefficient of kinetic friction μ k μ_k between the runner and the pavement is 0.1 0.1 and the angle φ φ between the cord and the pavement is 3 0 . 30^\circ. Find the magnitude of the tension force by which the boy affects the sled.


Details and assumptions:

  • Take g = 9.81 m/s 2 . g = 9.81 ~ \text{m/s}^2.
  • Give your answer in Newton to two decimal places.


The answer is 21.41.

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1 solution

Muhammad Erfan
Oct 15, 2017

the force diagram:

The velocity of the sledge is constant so the acceleration of the sledge equal to zero in the direction of x axis or F x = 0 \sum { F_{x} } =0

F x = T cos 30 ° f = 0 \sum { { F }_{ x } } =T\cos { 30° } -f=0

Normal Force acting on the sledge: N = m g T sin 30 ° N=mg-T\sin { 30° }

f = μ k N = μ k ( m g T sin 30 ° ) = T cos 30 ° f={ \mu }_{ k }N={ \mu }_{ k }\left( mg-T\sin { 30° } \right) =T\cos { 30° }

f = 0.1 [ ( 20 × 9.81 ) T sin 30 ° ] = T cos 30 ° f=0.1\left[ \left( 20\times 9.81 \right) -T\sin { 30° } \right] =T\cos { 30° }

f = 19.62 T ( 0.1 sin 30 ° ) = T cos 30 ° f=19.62-T\left( 0.1\sin { 30° } \right) =T\cos { 30° }

T = 19.62 ( 0.1 sin 30 ° + cos 30 ° ) 21.41 T=\frac { 19.62 }{ \left( 0.1\sin { 30°+\cos { 30° } } \right) } \approx \boxed{21.41}

so, the tension force by which the boy affects the sledge is 21.42 21.42 newton

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