(a) If half of the weight of a small 1.00×103 kg utility truck is supported by its two drive wheels, what is the magnitude of the maximum acceleration it can achieve on dry concrete? (b) Will a metal cabinet lying on the wooden bed of the truck slip if it accelerates at this rate? (c) Solve both problems assuming the truck has four-wheel drive.
Step-by-step solution Step 1 of 4 (a) The half of the weight of the truck supported by the driving wheels 1x10 N 3 So the normal reaction force from the ground is; So the friction force action on the driving wheel The acceleration of the truck is caused by this friction force. From the second law of Newton the acceleration of the truck is . Step 2 of 4 (b) When the truck accelerated forward, the metal cabinet that is loaded on to the truck experience backward force of inertia. If the backward force of inertia is bigger than friction force between the metal cabinet ant the wooden bed of truck. It starts to slide backward. Now the maximum acceleration possible without moving the cabinet is equal to the acceleration applied, so the cabinet will not move.