Question

Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angle of 6.7° with the horizontal. What is the magnitude Frope of the force on the skier from the rope when (a) the magnitude v of the skier's velocity is constant at 1.90 m/s and (b) v = 1.90 m/s as v increases at a rate of 0.150 m/s2?

Answer 1

a)
`F = 78.606\,N`, b)
`F = 88.911\,N`

Step-by-step explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

`\Sigma F_(x') = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_(y') = N - m\cdot g \cdot \cos \theta = 0`

The force on the skier is:

`F = m \cdot g \cdot \sin \theta`

`F = (68.7\,kg)\cdot (9.807\,(m)/(s^(2)) )\cdot \sin 6.7^(\textdegree)`

`F = 78.606\,N`

b) The equations of equilibrium are the following:

`\Sigma F_(x') = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_(y') = N - m\cdot g \cdot \cos \theta = 0`

The force on the skier is:

`F = m\cdot (a + g \cdot \sin \theta)`

`F = (68.7\,kg)\cdot (0.150\,(m)/(s^(2))+9.807\,(m)/(s^(2))\cdot \sin 6.7^(\textdegree))`

`F = 88.911\,N`