physics problem dealing with Friction?

Question

A sled weighing 60.0N is pulled horizontally across snow so that the coefficient of kinetic friction between the sled and the snow is 0.100. A penguin weighing 70.0N rides on the sled. If the coefficient of static friction between the penguin and the sled is 0.700, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.

I'm not sure what the answer is.

I asserted that since the penguin is on the sled the normal force of the system is 130N, so I found the force of friction using the coefficient of kinetic friction to be 13N. Then I found the force of friction of the penguin sliding off, using the coefficient of static friction and a normal force of 70N of the penguin, to be 49N. I said that since it is moving at a constant speed there has to be a force of 13N on the sled already, so if I subtract 13N from 49N I would get 36N. So 36N is the force that is needed to slide the penguin off, right? or would it be 13N+49N=62N? Thanks so much!

Answer 1

It doesn't matter that there is already a force on the sled since the question is to compute the maximum force before the penguin begins to slide. It only matters that the force is greater than the frictional force, otherwise the sled would come to rest and a greater force would have to be applied to overcome the static coefficient of friction of the sled and the snow.

You computed that correctly to be 49N on the penguin.
Since the force is applied to the sled it is the 49N + the frictional force of the sled on the snow, which is 13N. Any additional force will simply accelerate the sled and penguin until the penguin slips,
so the correct answer is 13+49=62N

j

Answer 2

i'm hoping this helps For this type of subject you would be able to desire to think approximately each and all the forces appearing on the vehicle, and use the physics of action in a circle to calculate the solutions you desire. The forces on the vehicle are:- its weight w, the traditional tension N, the frictional tension Fs the load balances the traditional tension it is N = w = mg The horizontal friction tension would desire to grant the centripetal tension performing in direction of the centre of the circle (of which the curve is an arc) making use of Newton's 2d regulation we've Fs = ma = mv^2/ r The friction is static with the aid of fact there is not any lateral slippage. on the utmost a possibility velocity the frictional tension has its maximum value for this reason Fs = Fsmax = us x mg (us right it is coefficient of friction) and for this reason us x mg = mv^2 / r Cancelling out the hundreds and taking the sq. roots of the two factors of the equation supplies sqrt (us x g x r) = v so max velocity v = sqrt (0.sixteen x 0.80 one x 75 ) = 10.80 5 m/s At this velocity the vehicle is in basic terms close to slipping..... Any swifter, disaster ! you will locate that the utmost velocity does not remember on the mass of the vehicle, yet in basic terms on the radius of the curve and the cost of the coefficient of friction. stable success with the examination