Question

A 16.0-m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 57.0° angle with the horizontal.

(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.20 m along the ladder from the bottom.

Horizontal Force

magnitude | |

direction |

Vertical Force

magnitude | |

direction |

(b) If the ladder is just on the verge of slipping when the
firefighter is 9.10 m from the bottom, what is the coefficient of
static friction between ladder and ground?

Answer #1

a)

Apply equiliberium condition,

N2(16*sin57) = 850*(4.20*cos57) + 490*(16/2)cos57

N2 = [850*(4.20*cos57) + 490*(16/2)cos57] / (16*sin57)

N2 = 304 N

Horizontal component of the force is

N2 = 304 N directed right

verticle component of the force is

N1 = 490 N + 850 N = 1340 N directed upward

b)

static friction when fighter is at 9.10 m

N2(16*sin57) = 850*(9.10*cos57) + 490*(16/2)cos57

N2 = [850*(9.10*cos57) + 490*(16/2)cos57] / (16*sin57)

N2 = 473.1 N

coefficient of static friction is

= 473.1 N / 1340 = 0.353

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