Question

A rectangular block that is 0.6m wide, 0.6m long and 1.2m high stands on a horizontal loading plane of a car. The block has uniform mass distribution and the total mass is 80kg. The block is oriented with the sides parallel to the loading plane. The static friction coefficient between the block and the loading plane is 0.6. The car’s speed is 60 km/h, when the driver detects a warning signal and brakes sharply, but evenly. a) What is the shortest stop time the car can have without the block slipping on the loading plane? b) Draw a picture of the situation and indicate the direction of the relevant forces when the car brakes. c) What is the shortest stop time the car can have without the block overturning on the loading plane?

Please explain briefly and step by step.

Answer 1

a) When can decelerates formed at due to inertia block will try to remain at rest, It is the friction which will decelerate the block to the more along with car. If frictional force is less than the force on block due to deceleration of car Then the block will slip. Hence fictional force on the block must be equal to or greater than the decelerating force on it. → Ma = ung وتر = a a=0.689.8m/s² [a=5.88 m/s2 Kinematic equation for car Uf = vi + aut Y = 0, y = 60 km/h = 100 mis a = - 5.88 m/52 (siegative since braking? 50 = 100 - 5.88 + =+ = 5.675 suppose the car is moving to right, braking force will act to left on it. If we consider deceleration

a frome attached to car the forces acting of block will be as → motion FBD Block FN Bradering > of TA→ Ma force L aar Img force fu normal For equilibrium F - Mg = 0 f frictional force * Ę - Ma = 0 Mg weight (since frame is attached to car (non-inestial frame therefore Ma is to right) c) Toppling will occur due to unbalanced torque To avoid toppling torque due to ma must be countered by torque due to mg. On the verge of toppling on centero B will be in contact with the plane of car. Hence normal force ass well as frictional force hare line of actions through B Moment about B 0.3 mg - 0.6 Mа = 0 → a= 4.9 m/s2 ts = 6.8087 ma 03m mg 6.80