How Much Torque Does a Person Exert on a Ladder?

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To calculate the torque exerted by a person standing on a ladder leaning against a house at a 20-degree angle, the formula T = rFsin(x) is used, where r is the distance from the pivot point, F is the weight of the person, and x is the angle. With a weight of 650 N and a distance of 1.0 m, the torque is calculated to be approximately 222.3 J, directed counter-clockwise when viewed from the house's perspective. The right-hand rule confirms the direction of torque, which is out of the plane towards the observer. If the ladder were positioned on the opposite side, the torque would still be 222.3 J but would be considered negative due to the clockwise direction. Understanding the signs and directions of torque is crucial for consistent calculations in physics.
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Imagine that a ladder leans against the house at an angle of 20o with respect to the vertical. A person with a weight of 650 N stands on the ladder at a point 1.0 m from its bottom. What are the magnitude and direction of the torque that this person exerts on the ladder around the point where the ladder touches the ground?

Where do I start?
 
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courtney1121 said:
Imagine that a ladder leans against the house at an angle of 20o with respect to the vertical. A person with a weight of 650 N stands on the ladder at a point 1.0 m from its bottom. What are the magnitude and direction of the torque that this person exerts on the ladder around the point where the ladder touches the ground?

Where do I start?
Start by drawing a sketch, showing the angle the ladder makes with the wall, then look up the definition of torque. What do you get?
 
Well torque will be T=rFsinx

x = 20 degrees
r = 1m
F=650

so T=1(650)sin(20) = 222.3?
 
Is the direction towards me then? If I am looking at the house, the ladder is on the right of the house.
 
From using the right-hand rule, I determined that the torque will be counter-clockwise towards me...but I usually mess up the right hand rule somehow...
 
courtney1121 said:
Is the direction towards me then? If I am looking at the house, the ladder is on the right of the house.
Magnitude is correct, don't forget units!(joules).
Direction is also correct, using right hand rule, curl your fingers of your right hand in the direction of the force, with your thumb in the direction of the axis about which the torque is taken. Your thumb will point in the direction of the torque, which in this case is out of the plane towrd you along the z axis. The torque is counterclockwise. For practice, try the same problem, but with the ladder on the left side of the house. What do you get for a magnitude and direction of the torque in this case?
 
well, the torque will be the same but shouldn't it be negative so -222.3J and the direction would be towards me again but clockwise?
 
courtney1121 said:
well, the torque will be the same but shouldn't it be negative so -222.3J and the direction would be towards me again but clockwise?
What are you afraid of straining your wrist? The direction is clockwise, but AWAY from you this time. You must get that thumb passing through the point about which the torque is taken. That negative sign indicates that the direction is opposite the direction in the first case when the ladder was on the other side. Usually clockwise moments are considered positive, and counter clockwise moments are negative, but it doesn't generally matter too much as long as you are consistent. Try it again, and strain your wrist and you'll get it!
 
haha people are looking at me like I'm trying to hitch hike in my room...but i see now why it is pointing away from me.
 
  • #10
courtney1121 said:
haha people are looking at me like I'm trying to hitch hike in my room...but i see now why it is pointing away from me.
Ya, thanks a lot, now i need an Ace bandage.
 
  • #11
im sorrrry! :(
 
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