Students picking apples, 40* ladder on fence

[clos561]

Homework Statement

Two kids are trying to pick some apples that are on the other side of the fence which is 1.5m high. Since there is a vicious dog on the other side, they decided to use a ladder and lean it against the fence. The ladder of negligible mass is placed at an angle of 40° to the horizontal. One kid of mass 50 kg stands at the bottom of the ladder while another kid of mass 40 kg goes up the ladder. How far up along the ladder can the lighter kid go before it tips over?

Homework Equations

dont know if this is correct. Mass1Radius1=Mass2Radius2 could not find examples in my textbook that are similar

The Attempt at a Solution

the answer i got was 2.91 meters but its not an option.
The book says for a system like this to be equilibrium that equation must be true. not sure how to approach this problem.

 

[RoshanBBQ]

Homework Statement

Two kids are trying to pick some apples that are on the other side of the fence which is 1.5m high. Since there is a vicious dog on the other side, they decided to use a ladder and lean it against the fence. The ladder of negligible mass is placed at an angle of 40° to the horizontal. One kid of mass 50 kg stands at the bottom of the ladder while another kid of mass 40 kg goes up the ladder. How far up along the ladder can the lighter kid go before it tips over?

Homework Equations

dont know if this is correct. Mass1Radius1=Mass2Radius2 could not find examples in my textbook that are similar

The Attempt at a Solution

the answer i got was 2.91 meters but its not an option.
The book says for a system like this to be equilibrium that equation must be true. not sure how to approach this problem.

It would be nice if you showed your work. It seems you did this:
$$\frac{1.5}{sin(\frac{40}{180}\pi)}\frac{50}{40} = 2.92$$

Is that what you did? Were you balancing the torque about the point in contact with the fence?

 

[clos561]

yes that is what i did.

 

[RoshanBBQ]

yes that is what i did.

The distance up the ladder is equal to that quantity plus the distance to the pivot. Add 1.5/sin(40 degrees) to your answer.

 

[asteriatic]

I would first clarify the problem statement to make sure that all the given information is accurate and complete. I would also ask for the units of measurement to ensure consistency in calculations.

Next, I would draw a free body diagram of the system to understand the forces acting on it. This would include the weight of both kids, the weight of the ladder, and the reaction forces from the fence and the ground. I would also consider the center of mass of the system and how it shifts as the lighter kid moves up the ladder.

From the given information, it seems that the ladder is considered to be massless, but in reality, it would have some mass and this would affect the stability of the system. Therefore, I would consider the mass of the ladder in my calculations.

To solve for the distance the lighter kid can go up the ladder before it tips over, I would use the principles of rotational equilibrium. This would involve setting up an equation that equates the torque (due to the weight of the kids and the ladder) on one side of the system to the torque on the other side (due to the reaction forces from the fence and the ground). I would also consider the angle at which the ladder is placed and the distances of the forces from the pivot point (the point where the ladder meets the ground).

Once I have set up the equation, I would solve for the unknown distance using basic algebraic manipulations. I would also make sure to include the units in my calculations and convert them if necessary.

In conclusion, as a scientist, I would approach this problem by first clarifying the given information, drawing a free body diagram, considering the mass of the ladder, and using principles of rotational equilibrium to solve for the unknown distance. I would also make sure to include units in my calculations and convert them if necessary.