Consider the ladder in the figure. Assume there is friction between the verticle wall and the ladder, with the coeffecient of static friction=0.30. Find the angle at which the ladder just begins to slip.
Suppose a crate of mass 7.5 kg is placed on the plank in the figure below, at a distance of 3.9m from the left end. Find the forces exerted by the 2 supports on the plank.
The Attempt at a Solution
I've used the tourque...
A game uses a spring launcher to shoot a puck with mass of 0.5kg along a frictionless track as shown in the diagram. The spring has a k value of 600 N/m. By how much does the spring need to be compressed to launch the puck and have it go around the inside of the loop...
An 80 kg stuntman is shot from a cannon at a circus. The cannon is 3m in length and angled at 35 degrees above the level floor. The landing cusion is 30m away from the cannon. What velocity will the stuntman have 1)when he leaves the cannon and 2)when he first hits the...
Consider the problem of a car traveling along a banked turn. Sometimes roads have a "reversed" banking angle. That is the road is tilted "away" from the center of curvature of the road. The coefficient of static friction between the tires and the road is \mus=0.50, the...
Jeff Gordon (a race-car driver) discovers that he can accelerate at 4.0m/s2 without spinning his tires, but if he tries to accelerate more rapidly he always "burns rubber."
Find the coefficient of friction between his tires and the road. Assume the force from the engine is...
Okay so the forces acting on the small crate are a frictional force in the negative x direction wich is what I calculated out to be be 75.537. There is also a force from the larger mass pushing into the smaller crate... and is there also the force of the crates' velocities?
Alright, I did the math based on the small great, and I got the right answer (basically)
Ffriction=the coefficient of friction*the normal force.
That's a lot better. I don't understand why you use the small crate to solve this problem though.
Your moving company runs out of rope and hand trucks, so you are forced to push two crates along the floor. The crates are moveing at constant velocity, their masses are m1=45kg and m2=22kg, and the coefficient of kinetic friction between both crates and the floor are 0.35...
So, basically that means the total force acting on the car is the velocity and the frictional force. After ignoring the mass though, I'm still stuck. I don't know how to find the frictional force or the coefficent of friciton when ignoring mass.
A car is moving with a velocity of 20m/s when the brakes are applied and the wheels lock (stop spinning), The car then slides to a stop in 40m. FInd the coefficient of kinitic friction between the tires and the road.