Question

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 810-N firefighter has climbed 3.90 m along the ladder from the bottom. horizontal force magnitude N direction vertical force magnitude N direction (b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground? (c) What If? If oil is spilled on the ground, causing the coefficient of static friction to drop to half the value found in part (b), what is the maximum distance (in m) the firefighter can climb along the ladder from the bottom before the ladder slips? m

Answer 1

Answer:

A) Horizontal force is 329.55 N and it's direction is towards the wall while vertical force is 1300 N and it's direction is upwards.

B) coefficient of friction = 0.425

C) maximum distance = 2.58m

Explanation:

I have attached a a drawn free body diagram for clarity.

A) First of all let's get the sum of moments about the floor contact to find the wall reaction horizontal force.

Thus, summation of horizontal forces gives;

Rw[14sin55] - 490[(14/2)cos55] - 810[3.9cos55] = 0

11.468Rw - 1967.37 - 1811.93 = 0

11.468Rw - 2991.35 = 0

Rw = 3779.3/11.468

Rw = 329.55 N and it's direction is towards the wall

Now, let's get the summation of the vertical forces to zero to find the floor vertical force

Thus;

Fv - 490 - 810 = 0

Fv = 490 + 810 = 1300N

Fv = 1300 N upward

B) Now the firefighter is higher i.e 9.40 m from the bottom.

Using the same procedure in part a to find the horizontal reactions, we'll obtain;

Rw[14sin55] - 490[(14/2)cos55] - 810[9.4cos55] = 0

11.468Rw - 1967.37 - 4367.21 = 0

11.468Rw - 6334.58 = 0

11.468Rw = 6334.58

Rw = 6334.58/11.468

Rw = Fh = 552.37 N

The vertical reaction remains the same as

Fv = 1300 N upward

Coefficient of friction μ is given as the ratio of the maximum horizontal force to vertical force

Thus,

μ = Fh / Fv

μ = 552.37/1300

μ = 0.425

C) oil spilled and caused the coefficient of friction to be half its value. Thus, μ = 0.425/2 = 0.2125

Now, taking moments about the base of the ladder, we have;

Let the maximum distance the firefighter can climb be "d" from the horizontal. Thus, we now have;

(1300x0.2125x14Sin55) - (490(14/2)Cos55) - (810xdxCos55)= 0

3168.07 - 1967.37 - 464.6d = 0

1200.7 = 464.6d

d = 1200.7/464.6

d = 2.58m

Answer 2

Answer:

Check attachment for the free body diagram

Explanation:

Given that,

Ladder length L = 14m

Weight of ladder W= 490N

The weight will act at the midpoint

i.e at 14/2 = 7m, L1 = 7m

The ladder makes an angle of 55° with the horizontal. θ=55°

Weight of firefighter Wf =810N

The firefighter is at 3.9m from the horizontal ground, L2 =3.9

The wall exerts a force on the ladder, let It be Nw

The ground exerts a force on the ladder, let it be Ng

The let Ff be the frictional force that opposes motion.

a. We want to find the horizontal and vertical force the ground exerted on the ladder i.e Ng and Ff.

Using Newton second law

ΣFy= m•ay

ay=0, since the body is not accelerating

ΣFy = 0

Ng — Wf —W = 0

Ng = Wf + W

Ng = 490 + 810

Ng = 1300 N.

Also,

ΣFx= m•ax

ax=0, since the body is not accelerating

ΣFx = 0

Ff — Nw = 0

Ff = Nw

Now,

Let take moment about point A(ground), but note before we take moment, the forces must be perpendicular to the ladder.

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•3.9 + (490Cos55)•7 = (NwSin55)•14

1811.93 + 1967.37 = 11.47Nw

3779.3 = 11.47Nw

Then, Nw = 3779.3/11.47

Nw= 329.49N

Since Nw = Ff

Then Ff = 329.49N

So the required reaction exerted by the ground on the ladder are

Ng = 1300 N

Ff = 329.49 N

b. Now the firefighter is a distance of 9.4m from the horizontal and the ladder is about to slip

So we need to calculate the coefficient of static friction μs

Check attachment for new diagram,

So calculating for Nw again, since the firefighter have new position

Now the firefighter is at 9.4m from the ground. Therefore, L2 = 9.4m

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•9.4 + (490Cos55)•7 = (NwSin55)•14

4367.21 + 1967.37 = 11.47Nw

6334.58 = 11.47Nw

Then, Nw = 6334.58/11.47

Nw= 552.274N

Since Nw = Ff

Then Ff = 552.274N

Then, using frictional law

Ff = μs•Ng

The Ng doesn't change

Then, 552.274 = 1300μs

μs = 552.274/1300

μs = 0.42

c. Now, we want to know the maximum distance of the firefighter if the coefficient of static friction is reduce by half

Then, μs = 0.42/2

μs = 0.21

The assume the firefighter is at L2 from the horizontal.

Then, the frictional force is

Ff = μsNg

Ff = 0.21 × 1300

Ff = 273N

Then, Nw = Ff = 273N

Taking moment about point A

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•L2 + (490Cos55)•7 = (273Sin55)•14

464.6•L2 + 1967.37 = 3130.8

464.6•L2 = 3130.8—1967.37

464.6L2 = 1163.43

L2 = 1163.43/464.6

L2 = 2.5m

The maximum distance before the ladder begin to slip is 2.5m.