Friction — Question 7

Section 6-1 · Physics pages 784–785

Question 7 Static friction can transmit a pull to another block Take g = 10 m/s2. A 2 kg block rests atop a 3 kg block. The lower block sits on a frictionless floor and is pulled right with 10 N. The upper contact can supply at most 5 N of static friction. Initially both blocks are at rest. 3 kg 2 kg 10 N Only the floor contact is frictionless. (a) If they stick, find acceleration and contact friction. (b) Identify the reaction to the friction force on the upper block. (c) Find the largest pull that permits sticking.
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Question 7 - Solution Static friction can transmit a pull to another block 3 kg 2 kg 10 N (a) Use the pair and then the upper block. For the two-block system, internal contact forces cancel: a = P mu + ml = 10 2 + 3 = 2 m/s2. The upper block’s horizontal equation then gives f = mua = 2(2) = 4 N rightward < 5 N. (b) Keep the two recipients distinct. Newton’s third law acts between different bodies:⃗ fl→u = +4ˆi N,⃗ fu→l = −4ˆi N . The lower block confirms the same acceleration: 10 −4 = 3(2) N. (c) Apply the upper-block bound. For a general nonnegative pull, a = P 5 , f = 2a = 2P 5 ≤5 N. Pmax = 12.5 N . The limiting constraint is the upper contact, although the floor itself is frictionless.

Question 7

Static friction can transmit a pull to another block

Take g=10g=10 m/s2^2. A 2 kg block rests atop a 3 kg block. The lower block sits on a frictionless floor and is pulled right with 10 N. The upper contact can supply at most 5 N of static friction. Initially both blocks are at rest.

Diagram shown in the original SVG worksheet.

  1. If they stick, find acceleration and contact friction.

  2. Identify the reaction to the friction force on the upper block.

  3. Find the largest pull that permits sticking.

Question 7 - Solution

Static friction can transmit a pull to another block

Diagram shown in the original SVG worksheet.

(a) Use the pair and then the upper block.

For the two-block system, internal contact forces cancel: a=Pmu+ml=102+3=2m/s2.a=\frac{P}{m_u+m_l}=\frac{10}{2+3}=2\ \mathrm{m/s^2}. The upper block’s horizontal equation then gives f=mua=2(2)=4N rightward<5N.\boxed{f= m_u a=2(2)=4\ \mathrm N\text{ rightward}}<5\ \mathrm N.

(b) Keep the two recipients distinct.

Newton’s third law acts between different bodies: f→l→u=+4𝒊̂N,f→u→l=−4𝒊̂N.\vec f_{l\to u}=+4\hat{\mathbf i}\ \mathrm N,\qquad\boxed{\vec f_{u\to l}=-4\hat{\mathbf i}\ \mathrm N}. The lower block confirms the same acceleration: 10−4=3(2)10-4=3(2) N.

(c) Apply the upper-block bound.

For a general nonnegative pull, a=P5,f=2a=2P5≤5N.a=\frac P5,\qquad f=2a=\frac{2P}{5}\leq5\ \mathrm N. Pmax=12.5N.\boxed{P_{\max}=12.5\ \mathrm N}. The limiting constraint is the upper contact, although the floor itself is frictionless.