8/14 points |Previous My Note 16.0 m uniform ladder weighing 510 N rests against

Question

8/14 points |Previous My Note 16.0 m uniform ladder weighing 510 N rests against a frictionless wall. The ladder makes a 56.0° 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 is 4.00 m from the bottom. Magnitude of the horizontal force Enter a number. Direction O away from the wall O towards the wall Magnitude of the vertical force 1370 NDirection O down O up (b) If the ladder is just on the verge of slipping when the firefighter is 8.90 m up, what is the coefficient of static friction between ladder and ground? Your response differs from the correct answer by more than 10%. Double check your calculations. Answer Save Progress My e a basket of food hanging at the end of the beam. The beam is hynary hear weighing 720 N walks out on a beam in an attempt to retriev

Explanation / Answer

Part A.

Using force balance

in x-direction

total Fx = 0, since it's not moving

0 = Ffriction - Nwall

Nwall = Ffriction

in y-direction

total force = 0

Nground - Wfire - Wladder = 0

Nground = 850 + 510 = 1360 N

using torque balance

torue in CCW = torque in CW

torque)wall = torque)ladder + torque)fire

Nw*cos 34 deg*16 = 850*4.0*sin 34 deg + 510*8*sin 34 deg

Nw = (850*4*sin 34 deg + 510*8*sin 34 deg)/(16*cos 34 deg)

Nw = 315.33 N to the left

Ff = 315.33 N to the right

Ng = 1360 N

B.

again same torque balance

torue in CCW = torque in CW

torque)wall = torque)ladder + torque)fire

Nw*cos 34 deg*16 = 850*8.90*sin 34 deg + 510*8*sin 34 deg

Nw = (850*8.90*sin 34 deg + 510*8*sin 34 deg)/(16*cos 34 deg)

Nw = 490.91 N

Nw = Ff

Ff = uk*Ng

uk*Ng = Nw

uk = 490.91/1360

uk = 0.361 = Coefficient of static firction

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