Question

A block of mass m₁ on a rough, horizontalsurface is connected to a ball of mass m₂ by alightweight cord over a lightweight,frictionless pulley, as shownin Figure 5.21a. A force of magnitude F at an angle ofθ with the horizontal is applied to the block asshown. Thecoefficient of kinetic friction between the block andthe surface is µ_k.


Figure 5.21a
The external force F applied as shown can causethe block to accelerate to the right.


Consider the system described in the example withm₁ = 0.40 kg and m₂ = 0.51kg. The coefficient of static friction between theblock and thesurface is 0.45. The angle θ of the force Fis equal to 28.0°. If F = 0,you can easily show that the block will accelerate tothe leftsince the maximum static friction force is not sufficient to keepthe block at rest. If F is sufficiently large, it is clearthat the block willaccelerate to the right. Find the range ofF that allows the system to remain at rest.

Answer 1

to keep from moving left, Forces acting on m1 = 0or m2*g = μ*m1*g + Fcos28F = g*(m2-μm1)/cos28 = 9.8(0.51-0.45*0.4)/cos28 = 3.66 N(to keep it from moving to left)Similarly to keep it from moving to right,m2*g+μm1*g = Fcos28solve for F, F = 9.8(0.51+0.45*0.4)/cos28 = 7.66 Nanswer: 3.66<= F<= 7.66 (N)