Recent content by Dannystu

Homework Statement A Pulley has mass M=4 kg and radius R=0.4 m. Assume that it is a uniform solid disk so that its moment of inertia is I= .5MR^2. A massless cord is wrapped around it and a tension force Ft is applied. The pulley starts from rest. After the tension force has been applied for...

Homework Statement It has been a while since I have taken pre-cal, thus I am not very good with trig. A man pushes a dolly with a force of 82 N which makes an angle of 35 degrees below the horizontal. Is the force along the x-axis F*sin(theta)? Or cos?

thanks a lot Doc!

Homework Statement A ball, mass m = 1.5 kg is twirled at constant speed v at the end of a massless string in a vertical circle of radius r = 1.2 m. The period for this uniform circular motion is T = 1.3 s. Free body diagrams for the ball at the top and at the bottom of the circle are shown...

anyone?

Yes because the car is not moving on the vertical direction, and the friction force in pointed in the opposite direction of motion.

Is the normal force equal and oppositely directed to the weight? I drew the FBD showing the a front view of the car.

Homework Statement A car, mass m=2500kg, rounds a curve on a flat road at a speed v= 15 m/s. The radius of curvature of the curve is r= 60m. There is obviously (static) friction between the road and the car tires, or the car would not stay on the curve. Homework Equations a.)...

Work-energy theorem: W=f*d*cos(theta) and W= KE final- KE initial so: (1/2)*3000*(26.19)^2- (1/2)* 3000*(20)^2 = 428,874J Is that right?

Okay thanks! Could you possibly push me in the right direction on how to approach part d?

Homework Statement The figure shows a roller coaster car mass m= 3,000kg, on a portion of a roller coaster ride. The height difference between points A and B is 35m. The height difference between B and C is 23m. The car starts from rest at point A. Take y=0 point to be point B. For parts...

Okay here goes my attempt: a.) The object does not move on the vertical axis, so Fn=m*g 9.8*10=98 Normal force acting on object one is 98N b.) For object one- The only force acting on the x-axis is the tension. So: T=m*a For object two- I came up with -m2g+T=m*a c.) So, For object one...

So the normal force is greater at the bottom? Correct? Thanks!

I drew my free body diagram- when the rider is on top, I have the normal force pointing up and weight and acceleration pointing down towards the center of the wheel. When the rider is at the bottom, the normal force and acceleration are pointing up and the weight is pointing down. For top-...

Homework Statement Two objects of masses, m1=10 kg and m2=20kg are connected by a cord over pulley. m1 sits on a horizontal frictionless table. The system is released and m1 accelerates to the right while m2 accelerates downward. Homework Equations a.) Calculate the magnitude of the...