Friction — Question 2

Section 6-1 · Physics pages 774–775

Question 2 A moving support supplies the required static friction Take g = 10 m/s2. A 3 kg crate rests on a truck bed. The truck accelerates right at 2 m/s2. The contact can supply up to 9 N of static friction; no other horizontal force acts on the crate. block atruck The crate and bed share acceleration only if the contact can supply the required force. (a) Find the actual friction if the crate remains fixed on the bed. (b) Find the largest truck acceleration magnitude compatible with sticking. (c) If the truck moves right but accelerates left at 1 m/s2, find friction and explain its direction.
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Question 2 - Solution A moving support supplies the required static friction block atruck (a) Demand common acceleration. Sticking requires the crate to share the truck acceleration: fs = ma = 3(2) = 6 N < 9 N. Thus fs = 6 N rightward is feasible; the actual force is below its maximum. (b) Apply the available-force bound. The available static-friction interval imposes |ma| ≤fs,max =⇒|a| ≤9 3 = 3 m/s2 . (c) Follow acceleration, not velocity. With right positive, the new acceleration is a = −1 m/s2: fs = ma = 3(−1) = −3 N . Friction acts left to slow the crate with the bed. Without it, the crate would slide forward relative to the decelerating truck.

Question 2

A moving support supplies the required static friction

Take g=10g=10 m/s2^2. A 3 kg crate rests on a truck bed. The truck accelerates right at 2 m/s2^2. The contact can supply up to 9 N of static friction; no other horizontal force acts on the crate.

Diagram shown in the original SVG worksheet.

  1. Find the actual friction if the crate remains fixed on the bed.

  2. Find the largest truck acceleration magnitude compatible with sticking.

  3. If the truck moves right but accelerates left at 1 m/s2^2, find friction and explain its direction.

Question 2 - Solution

A moving support supplies the required static friction

Diagram shown in the original SVG worksheet.

(a) Demand common acceleration.

Sticking requires the crate to share the truck acceleration: fs=ma=3(2)=6N<9N.f_s=ma=3(2)=6\ \mathrm N<9\ \mathrm N. Thus fs=6N rightward\boxed{f_s=6\ \mathrm N\text{ rightward}} is feasible; the actual force is below its maximum.

(b) Apply the available-force bound.

The available static-friction interval imposes |ma|≤fs,max⇒|a|≤93=3m/s2.|ma|\leq f_{s,\max}\Longrightarrow\boxed{|a|\leq\frac93=3\ \mathrm{m/s^2}}.

(c) Follow acceleration, not velocity.

With right positive, the new acceleration is a=−1a=-1 m/s2^2: fs=ma=3(−1)=−3N.\boxed{f_s=ma=3(-1)=-3\ \mathrm N}. Friction acts left to slow the crate with the bed. Without it, the crate would slide forward relative to the decelerating truck.