Vitani11 said:
Note that the tension force has a horizontal component to the right. What keeps the cylinder from sliding to the right?Vitani11 said:
Drawing the force as acting to the right only means that is taken as the positive direction for purposes of writing equations. It does not matter whether it will in fact act that way. If it doesn't, the value will come out negative. You just have to be consistent in the algebra.Vitani11 said:
Often a negative answer for an unknown force tells you that the direction of the force is opposite to what you assumed, but the magnitude is nevertheless correct.haruspex said:
Yes and no. In general, static friction force is constrained by ##|F_{sf}|<=\mu_s|F_{normal}|##. If told it is on the point of slipping that becomes ##|F_{sf}|=\mu_s|F_{normal}|##. The correct sign should emerge from consideration of the horizontal force balance.TSny said:
Yes. But if I blindly assume the wrong direction for the friction force, substitute |fs| = μs |N|, and then just mechanically go through the algebra, I get different results for the magnitudes of the forces compared to if I had assumed the correct direction for the friction force.haruspex said:
No, it's only a problem if you substitute fs = μs|N|.TSny said:
Well, maybe we can come back to this later. I don't want to write any equations at this point since Vitani11 has yet to show us any equations.haruspex said:
Your torque equation looks wrong in several ways. You are taking moments about the point of contact with the ground, right?Vitani11 said:
I don't know whether your φ is the 20 degrees or the 70 degrees, but either way, the horizontal distance from point of contact to mass centre is not (L/2)sinϕ, the horizontal distance to point of attachment of cable is not Lcos φ, and the vertical distance to point of attachment is not L sin φ. The cylinder has width.Vitani11 said:
However, ##\ \phi \ne 20^\circ \ ## .Vitani11 said:
But is not sufficient to change the angle. The length has to be changed too. I feel it is simpler not to worry about calculating the right length and the right angle, but simply work in terms of components of distance parallel to and normal to the cylinder's axis,SammyS said:
Tension refers to the force applied to an object in order to prevent it from moving or slipping. In the case of a cylinder about to slip, the tension in the support rope is what keeps the cylinder in place.
The tension in a support rope can be calculated using the formula T = mg + ma, where T is the tension, m is the mass of the cylinder, g is the acceleration due to gravity, and a is the angular acceleration of the cylinder. Alternatively, it can also be calculated using the formula T = F * sin(θ), where F is the force applied to the rope and θ is the angle of the support rope.
The tension in the support rope is affected by several factors, including the mass of the cylinder, the force applied to the rope, the angle of the support rope, and the coefficient of friction between the cylinder and its surface. Additionally, the length and elasticity of the rope can also affect the tension.
The optimal angle for the support rope will depend on the specific conditions of the cylinder and its surroundings. In general, a smaller angle will result in a higher tension in the rope, making it less likely for the cylinder to slip. However, the angle should also be practical and safe for the object and its surroundings.
The tension in the support rope can be adjusted by changing the force applied to the rope or the angle of the support rope. Increasing the force or decreasing the angle will result in a higher tension, while decreasing the force or increasing the angle will result in a lower tension. It is important to find the right balance to prevent the cylinder from slipping without causing any damage.
