static frictionkinetic friction

0503 Friction Forces

Kinetic and Static Friction

Exercises

(Mid Term 2) Block A, with a mass of 50kg, rests on a horizontal table top. The coefficient of static friction is 0.40. A horizontal string is attached to A and passed over a masseless, frictionless pulley as shown. The smallest mass mbm_b of block B, attached to the danging end, that will start A moving when it is attached to the other end of the string is:
Graph

Solution

mbg=μkmag⇒mb=μkma=20kg\begin{aligned} m_b g &=\mu_k m_a g\\ \To m_b &= \mu_k m_a = 20kg \end{aligned}

(Mid Term 2) A 40-N crate rest on a rough horizontal floor. A 12-N horizontal force is ten applied to it. If the coefficient of the friction are μs=0.5\mu_s = 0.5 and μk=0.4\mu_k = 0.4, the magnitude of the friction force on the crate is:

Solution

The maximum static friction force is 40×0.5=20N40 \times 0.5 = 20N. When an external force of 12N is applied upone the crate, it does not move, the static firction force equals to that external 12N force.

(Mid Term 2) A block is placed on a rough wooden plane. It is found when the plane it tilted 30°30\degree to the horizontal, the block will slide down at constant speed. The coefficient of kinetic friction of the block with the plane is:

Solution

Fs=μkmgcos⁡θmgsin⁡θ−Fs=0⇒μk=sin⁡θcos⁡θ=tan⁡30°\begin{aligned} F_s &= \mu_k mg\cos \th\\ mg\sin\th - F_s &= 0\\ \To \mu_k & = \frac{\sin\th}{\cos \th} = \tan 30 \degree \end{aligned}

(Mid Term 2) The system shown remains at rest. Each block weights 20N. The force of friction on the uppder block is:
Graph

Solution

The block is at rest. By Newton's Second Law

∑F=T−mgsin⁡θ−Fs=0=mg−mgsin⁡θ−Fs⇒Fs=8N\begin{aligned} \sum F &= T - mg\sin\th - F_s = 0\\ & = mg - mg\sin\th - F_s\\ \To F_s &= 8N \end{aligned}