Solve Physics HW: Kinetic Friction Coefficient

AI Thread Summary
To find the coefficient of kinetic friction for a 20kg sled being pulled with an 80N force at a 30-degree angle, the net force must be analyzed. The force of friction equals the pulling force's horizontal component, which can be calculated using trigonometry. The normal force is affected by both the weight of the sled and the vertical component of the pulling force. By applying the equation of motion and the definition of friction, the coefficient of kinetic friction can be derived. This problem emphasizes the importance of understanding forces and their components in physics.
jeffmazz
Messages
1
Reaction score
0
i feel really dumb but i have this physics homework question i can't get.

a 20kg sled is being pulled across a horizontal surface at constant velocity. the pulling force has a magnitude of 80N at a 30 degree angle above the horizontal. find the coefficient of kinetic friction.
 
Physics news on Phys.org
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top