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cmkc109
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Homework Statement
Torque = Fr sin theta, but is it always sin?
If I am looking for torque by gravity , will it by mg cos theta instead?
It's always sin(theta) when theta is defined as the angle from the radial direction of the applied force, so that theta = ±90° when the force is tangental (perpendicular to the radial direction).cmkc109 said:Torque = Fr sin theta, but is it always sin?
rcgldr said:It's always sin(theta) when theta is defined as the angle from the radial direction of the applied force, so that theta = ±90° when the force is tangental (perpendicular to the radial direction).
The problem states that the tension in the rope is perpendicular to the door, which would mean that theta as used in the torque equation is ±90°.
The angle of the trap door with respect to the floor (the angle from horizontal) is not the same as the angle used in your torque equation, torque = F r sin(theta). The angle of the trap door with respect to the floor is a different "theta".
It would probably be easier to state the angles relative to the trap door and the radius relative to the hinge of the trap door, so that you always use torque = force x radius x sin(angle of force). What is the direction of gravity relative to the trap door? What is the direction of the rope relative to the trap door?cmkc109 said:"with respect to the floor" do u mean the angle between the door and the floor (which is the force of gravity"?
That should be 90 - theta.cmkc109 said:sin ... cos so u don't need to convert the angle 180-theta
rcgldr said:It would probably be easier to state the angles relative to the door. What is the direction of gravity relative to the door? What is the direction of the rope relative to the door?
That should be 90 - theta.
You should be able to determine what the force from gravity is. Your class or textbook should have explained that you can assume that gravity acts as a downwards force applied to the middle of the trap door.cmkc109 said:Gravity is acting down, direction of the rope is acting upwards
(I don't know what the force is here, am i suppose to find that)
That should be T = mg x r/2 x cos(65°). 65° is the angle from the floor. The angle of gravity force from the door is 25°. If you used this, then the equation is T = mg x r/2 x sin(25°).cmkc109 said:I found the answer online
It uses cos65 x Fg x r/2
rcgldr said:That should be T = mg x r/2 x cos(65°). 65° is the angle from the floor. The angle of gravity force from the door is 25°. If you used this, then the equation is T = mg x r/2 x sin(25°).
rcgldr said:T = mg x r/2 x sin(25°).
Yes, the torques are equal in magnitude but opposing direction (equilibirum). The question is asking for the force in the rope. What is the equation for torque of the rope in terms of force from the rope?cmkc109 said:but the question is asking for the torque on the rope exerted on the door... torque of rope give sin90 = 1 ... is it because torque of rope = torque of gravity since it is an equilibrium torque?
rcgldr said:Yes, the torques are equal in magnitude but opposing direction (equilibirum). The question is asking for the force in the rope. What is the equation for torque of the rope in terms of force from the rope?
cmkc109 said:torque of rope give sin 90 = 1
rcgldr said:What is the equation for torque of the rope in terms of force from the rope?
You've already stated that theta for the rope is 90°. So for the rope, T = F r sin(90°), and for gravity T = mg x r/2 x sin(25°). You're also told that gravity = 9.8 m / s^2, and the mass of the trap door is 16.8 kg, so how many Newtons of force does gravity exert on the door? With this information, you should be able to solve for F, the force of the rope.cmkc109 said:T = rF sin theta
rcgldr said:You've already stated that theta for the rope is 90°. So for the rope, T = F r sin(90°), and for gravity T = mg x r/2 x sin(25°). You're also told that gravity = 9.8 m / s^2, and the mass of the trap door is 16.8 kg, so how many Newtons of force does gravity exert on the door? With this information, you should be able to solve for F, the force of the rope.
Yes, you stated that the torques were in equilibrium. Technically it's the sum of the torques that is zero, socmkc109 said:setting them equal to each other?
rcgldr said:Yes, you stated that the torques were in equilibrium. Technically it's the sum of the torques that is zero, so
Tgravity + Trope = 0
The force from gravity is negative, and the force from the rope is positive.
Torque caused by gravity is a rotational force that occurs when an object is acted upon by the gravitational force of another object. It is the product of the force of gravity and the distance between the two objects.
To calculate torque by gravity, you can use the formula: Torque = force x distance. The force is the gravitational force between the two objects, and the distance is the distance between their centers of mass.
The unit of measurement for torque by gravity is newton-meter (Nm) in the SI system. In other systems, it can also be expressed as pound-feet (lb-ft) or dyne-centimeter (dyn-cm).
The distance between objects has a direct impact on the magnitude of torque by gravity. As the distance increases, the torque by gravity decreases, and vice versa. This is known as the inverse square law.
Yes, torque by gravity can be negative. This occurs when the direction of the rotational force is opposite to the direction of motion. It means that the object is rotating in the opposite direction than expected.