course phy 121
I am having difficulty with the text problems #36-39 for chapter 4. Can you please check my work and questions and advise.#36 If the coeff of kin fric between a 35kg crate and the floor is 0.30, what horiz force is required to move the crate at a steady speed across the floor?
The four forces are gravity (35kg * 9.8m/s^s = 343N), normal force (343 N also) ( but sometimes in other problesm it isn't the same as the force of gravity and I don't understand why), kin frict (.30 * 343 = 102.9N) and the applied force (which has to be greater than or equal to kinetic fric.
What horiz force is required if coeff of kin fric is zero?
Since I multiplied 0.30 by 343, I think I just multiply 343 by zero and then the applied force would have to be greater than zero.
This is correct.
#37 A force of 48N is required tostart a 5.0kg box moving across a horz floor. what is the coeff of static frict? If the 48N force cont the box accel at .70m/s^2. What is the coeff of kin frict.
I got this one right according to the text and I feel somewhat comfortable with it.
#38 Suppose youare standing on a train accel at .20g. What minimum coeff of static frict must exist between your feet and thr floor if you are not to slide?
I started with mult .20 by 9.8 to get an accel of 1.96m/s^2. And I think the kinet force of frict multiplied by the normal force would be the force of frict, but I dont know how to figure that out and what to do with it.
To accelerate at 1.96 m/s^2 you would require a net force of 1.96 m/s^2 * M, where M is your mass.
The normal force between you and the floor would be equal to your weight, which is 9.8 m/s^2 * M.
The accelerating force must come from friction. Since
f_Frict < = coeff of frict. * normal force, we have
f_Frict / normal force < = coeff of frict, so
1.96 m/s^2 * M / (9.8 m/s^2 * M) < = coeff of friction so
.20 <= coeff of frict.
The coefficient of friction must be at least .20.
#39 What is the max accel a car can undergo if the coeff of static frict between the tires and the ground is 0.80?
I know coeff is .8 and I am trying to find accel. So I think I want to use a = Fnet / m. I can substitute Force applied - force of friction for Fnet. I can substitute coeff * normal force for force of friciton, but since I don't know where to go from there I assume I am on the wrong track.
You're doing OK here. The maximum possible static friction is f_Frict = .80 * normal force = .80 * 9.8m/s^2 * M, where M is the mass of the car.
a = f_Frict / M = .80 * 9.8m/s^2 * M / M = .80 * 9.8 m/s^2 = 7.8 m/s^2, approx..
Yes, I did read the chapter, but it was like reading greek. I understood pages 90, 91 and part of 92, but from pulling against friction on I got confused. Then I understood page 94 but got completely lost with the solution presented on page 95. I couldn't get past the fact that I though the x component is always magnitude * cos theta and the book has it as mag * sin theta. The equations confuse me with the symbols and moving things around and canceling, when I can't always figure out what was moved, substituted or canceled and why.
Thanks for any help you can offer." ""
The components of a weight parallel and perpendicular to an incline are indeed given by
wt_x = wt * cos(theta) and
wt_y = wt * sin(theta),
but only if theta is measured counterclockwise from the positive x-axis.
In the book's examples the angle of the incline is often given by theta. However this is not the angle of the weight with the positive x-axis, but rather the angle of the incline with respect to horizontal.