Anonymous

# Tension in rope and Max distance?

A 17.8 kg person climbs up a uniform 85.1 N ladder. The upper and lower ends of the ladder rest on frictionless surfaces. The bottom of the ladder is fastened to the wall by a horizontal rope that can support a maximum tension of 87.5 N. The angle between the horizontal and the ladder is 56 degrees. The acc. of gravity of 9.8 m/s^2

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a. Find the tension in the rope when the person is one-third of the way up the ladder; i.e., (10.2m/3 = 3.4 m). Answer in units of N.

b. Find the maximum distance Dmax the person can climb up the ladder before the rope breaks. Answer in units of m.

Relevance

Let's start by summing torques at the top of the ladder

(85.1*5.1+(10.2-D)*17.8*9.8)*cos(56)+

T*sin(56)*10.2=R*cos(56)*10.2

where D is the distance from the bottom of the ladder of our climber and

R is the reaction force at the foot, which is vertical

Summing vertical forces

R=85.1+17.8*9.8

T=28.7+D*11.5

so

a) When D=3.4

T=67.9 N

b)

Dmax

(87.5-28.7)/11.5=Dmax

Dmax=5.11 m

j

• A. The first thing to do is construct a free body diagram of the ladder. This will show you the forces acting on the ladder and the angles at which they act on the ladder. There are 5 forces acting on the ladder.

1)Reaction force of the wall at the top.

2)Reaction force of the floor at the bottom.

3)The Tension in the rope.

4)The force of gravity on the ladder.

5)The force of gravity on the person(which is in turn on the ladder)

Once you have the free body diagram, You can sum the forces in the x and y directions and sum the Moments of Forces or torque caused by the forces about any point on the ladder. Since the ladder is not accelerating, the forces in any direction and the moments will all sum to zero. This will give you 3 equations with 3 unknown variables and you can solve for all of them.

T = tension in the rope

Rb = reaction at top

Ra = reaction at bottom

Forces in the x direction

0 = T - Rb

Forces in the y direction

0 = Ra -17.8 - 85.1

Moment of force or torque = r x f (cross product) = rfsin(theta)

where theta is the angle between the ladder and the force.

0 =

T(5.1)sin56 + Rb(5.1)sin56 +17.8(1.8)sin34 - Ra(5.1)sin34

I just did algebraic substitution and I got

T = 32.58 N

You might want to check my calculations and make your own freebody diagram to make sure the angles I used are correct.

Part B.

Using the formulas from above you can see that Rb will always be equal to T and that Ra is constant.

When the person goes higher on the ladder the The moments due to the Rb, T, and the person will change. Rb and T are the dependent variables and the distance x of the person up the ladder is the independant variable. So using the equation for the sum of moments about the center of the ladder and knowin that Ra is always 102.9 N up, you can set up an equation like this.

0 = T(5.1)sin56 + Rb(5.1)sin56+17.8(5.1-x)sin34 - 102.9(5.1)sin34

Knowing that the rope will break at T = 87.5 N , you can substitute 87.5 for both T and Tb in the equation and solve for x ( distance up the ladder)

One tricky thing about the "distance up the ladder" part of the problem is whether or not they are asking the persons height above the ground or the distance from the person to the base of the ladder. I'll let you figure that part out.

I hope this helps and I hope all my equations were correct, again it is definately a good idea to draw your own free body diagram and make sure all my angles were correct.