Student on Ladder (Equilibrium)

[SuperCass]

Homework Statement

A student is standing on a ladder as shown in the figure below Each leg of the ladder is 2.7 m long and is hidged at point C. The tie-rod (BD) attached halfway up and is 0.76 m long. The student is standing at a spot 2.025 m along the leg and her weight is 478 N. (You may ignore the weight of the ladder and any minor friction between the floor and the legs.)

a) What is the tension in the tie-rod?

b) What is the vertical component of the force of the ground on the ladder leg at point A?

c) What is the horizontal component of the force of the ground on the ladder leg at point A?

d) What is the vertical component of the force of the ground on the ladder leg at point E?

Homework Equations

F=ma
Fg=mg
c^2 = a^2 + b^2 -2abcos(angle c)

The Attempt at a Solution

I used law of cosines to find the angles in the diagram.
I tried finding torques but I'm pretty sure I did those completely wrong.

Help me out please! I'm stuck!

 

[inky]

Homework Statement

A student is standing on a ladder as shown in the figure below Each leg of the ladder is 2.7 m long and is hidged at point C. The tie-rod (BD) attached halfway up and is 0.76 m long. The student is standing at a spot 2.025 m along the leg and her weight is 478 N. (You may ignore the weight of the ladder and any minor friction between the floor and the legs.)

a) What is the tension in the tie-rod?

b) What is the vertical component of the force of the ground on the ladder leg at point A?

c) What is the horizontal component of the force of the ground on the ladder leg at point A?

d) What is the vertical component of the force of the ground on the ladder leg at point E?

Homework Equations

F=ma
Fg=mg
c^2 = a^2 + b^2 -2abcos(angle c)

The Attempt at a Solution

I used law of cosines to find the angles in the diagram.
I tried finding torques but I'm pretty sure I did those completely wrong.

Help me out please! I'm stuck!

Consider two separated rods. How many forces exerted on each rod?
Use summation F(x)=0
summation F(y)=0
summation torque (C)=0

 

[SuperCass]

For the forces acting on the feet of the ladder, would that just be the normal force since it said to ignore friction? Are there others?

 

[hikaru1221]

No, there is only normal force, since friction is negligible as stated in the problem.

 

[SuperCass]

What other forces are in the x direction then other than the tension?
Would I do two normal forces for each leg?

 

[hikaru1221]

Well, in the x direction, there are only tensions and horizontal components of forces at C. However the forces at C are a bit complicated, because they also have vertical components.

All you have to do is to write down the equilibrium equations as inky pointed out:
- For the left leg, total F(x) = 0, total F(y) = 0, and total torque about any axis = 0.
- Similar things for the right leg.
So: What are the forces on the left leg? On the right leg?

 

[inky]

What other forces are in the x direction then other than the tension?
Would I do two normal forces for each leg?

(1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on let leg.

(2) Find sin (theta) and cos (theta) from right leg.

(3) Use summation F(y)=0 and summation torque(C)=0

(4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on let leg.

(5) Use summation F(y)=0 and summation torque(C)=0

(6)Now you got 4 equations. Find normal force at A and E.

(7) Find tension.

 

[inky]

(1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on let leg.

(2) Find sin (theta) and cos (theta) from right leg.

(3) Use summation F(y)=0 and summation torque(C)=0

(4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on let leg.

(5) Use summation F(y)=0 and summation torque(C)=0

(6)Now you got 4 equations. Find normal force at A and E.

(7) Find tension.

sorry . Typing error for no.(1) and no.(4)

(1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on right leg or not.

(4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on left leg or not.

 

[santi_h87]

sorry . Typing error for no.(1) and no.(4)

(1)Firstly draw right leg. How many forces acting on right leg? If we neglect friction ,check do you get 4 forces exerted on right leg or not.

(4) Draw left leg. How many forces acting on left leg? If we neglect friction ,check do you get 5 forces exerted on left leg or not.

inky, what are those 4 forces exerted on right leg? I got 3
1 is on C from the left leg, which has negative Y component and positive X component
2 is on E from the ground, which only has a positive Y component
3 would be the tension

the same for the left leg, I got 4 forces
1 is on C from the right leg, with a positive Y component and negative X component
2 is where the guy is standing, with only a negative Y component
3 would be the tension, the same as before
4 is the normal force from the ground to the left leg with only a positive Y component

 

[inky]

inky, what are those 4 forces exerted on right leg? I got 3
1 is on C from the left leg, which has negative Y component and positive X component
2 is on E from the ground, which only has a positive Y component
3 would be the tension

the same for the left leg, I got 4 forces
1 is on C from the right leg, with a positive Y component and negative X component
2 is where the guy is standing, with only a negative Y component
3 would be the tension, the same as before
4 is the normal force from the ground to the left leg with only a positive Y component

Hi santi_h87,
You are right as well. I count two forces at point C.