Skier velocity and friction problem

  • Thread starter Thread starter rcmango
  • Start date Start date
  • Tags Tags
    Friction Velocity
AI Thread Summary
The problem involves a skier being pulled up a 30° slope at a constant velocity, with a mass of 53.0 kg and a coefficient of kinetic friction of 0.190. To find the force exerted by the tow bar, it's essential to analyze the forces acting on the skier, including gravitational force, normal force, and frictional force. Since the skier moves at constant velocity, the net force is zero, allowing for the application of Newton's Second Law. A force-body diagram can help visualize the forces involved, and using trigonometric components will clarify the calculations. Understanding the role of friction is crucial for determining the total force exerted by the tow bar.
rcmango
Messages
232
Reaction score
0

Homework Statement



a skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 30.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 53.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.190. Find the magnitude of the force that the tow bar exerts on the skier.

Homework Equations





The Attempt at a Solution



since i know the slope is the acceleration i believe.
couldn't i just solve for acceleration, and maybe get more parts of the picture through trig from vector components?

I'm not sure how to use the friction though.
is there an equation that will help?
 
Physics news on Phys.org
"a skier is pulled up a slope at a constant velocity"

draw a force-body diagram and evaluate your forces and how you can solve for F.

visit my blog if you need examples.
 
Last edited:
Begin by defining a coordinate system (where you would like to place the x-axis), and tell me the forces that act in the x and y directions by using Newton's Second Law.
 
What rocophysics means is that since there is constant velocity, all forces should add to zero.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top