Question 1, chap 111, sect 1 part 1 of 2 10 points A plank having mass 7.1 kg ri

Question

Question 1, chap 111, sect 1 part 1 of 2 10 points A plank having mass 7.1 kg rides on top of two identical solid cylindrical rollers each having radius 5.3 em and mass 2.6 kg. The plank is puled by a constant horizontal force 6 N applied to its end and perpen- dicular to the axes of the cylinders (which are parallel)· The cylinders roll without slipping on a flat surface. There is also no slipping between cylinders and plank Find the acceleration of the plank Answer in units of m/s Question 2, chap 111, sect 1. part 2 of 210 points Find the frictional force acting on the plank. Answer in units of N.

Explanation / Answer

a) Newton's Second Law => acceleration = force/inertia

for the cylinders => I = 1/2mr^2

a = horizontal force/ (mass plank)+2(1/2mr^2) = 6N/(7.1kg)+(2.6kg*0.053^2)

= 0.85 m/s^2


b) the frictional force on the plank will be equal to the torque on each of the drums/radius...

so...T = rF = I*angular acceleration = 1/2*m*a = 1/2*2.6kg*0.85m/s^2

= 1.105 N of frictional force per cylinder or 2.21 N frictional force on the plank from both drums

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