In Fig. 6-34, blocks A and B have weights of 44 N and 22 N, respectively. (a) Determine the mi

In Fig. 6-34, blocks A and B have weights of 44 N and 22...

In Fig. $6-34,$ blocks $A$ and $B$ have weights of $44 \mathrm{~N}$ and $22 \mathrm{~N},$ respectively. (a) Determine the minimum weight of block $C$ to keep $A$ from sliding if $\mu_{s}$ between $A$ and the table is $0.20 .$ (b) Block $C$ suddenly is lifted off $A .$ What is the acceleration of block $A$ if $\mu_{k}$ between $A$ and the table is $0.15 ?$

In Fig. $6-34,$ blocks $A$ and $B$ have weights of $44 \mathrm{~N}$ and $22 \mathrm{~N},$ respectively. (a) Determine the minimum weight of block $C$ to keep $A$ from sliding if $\mu_{s}$ between $A$ and the table is $0.20 .$
(b) Block $C$ suddenly is lifted off $A .$ What is the acceleration of block $A$ if $\mu_{k}$ between $A$ and the table is $0.15 ?$

Key Concepts

Static Friction

Static friction is the force that resists the initiation of sliding motion between two surfaces in contact. It must be overcome by an applied force in order for motion to start, and its maximum value is typically given by the product of the coefficient of static friction and the normal force. This concept is crucial in scenarios where the aim is to prevent relative motion until a threshold is reached.

Kinetic Friction

Kinetic friction is the force that opposes the motion when two surfaces are sliding past each other. Unlike static friction, its magnitude remains constant (often given by the coefficient of kinetic friction multiplied by the normal force) once motion has commenced. Understanding kinetic friction is essential for analyzing systems after an object has overcome static friction and is in motion.

Normal Force

The normal force is the perpendicular contact force exerted by a surface on an object. It plays a critical role in calculating frictional forces, as both static and kinetic friction depend directly on the magnitude of the normal force. In many problems, accurately determining the normal force is key to finding the frictional force acting on an object.

Newton's Second Law

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This law is fundamental in dynamics, allowing the determination of acceleration given the net force, or vice versa, and is widely used in problems involving forces, masses, and accelerations.

Free-Body Diagrams

A free-body diagram is a graphical representation that depicts all the forces acting on an object. It is an essential tool in the analysis of mechanical systems because it helps isolate the object and clarify the direction and magnitude of forces, making it easier to apply Newton’s laws to solve for unknown quantities.

Transcript

00:01 Blocks a and b and blocks a is 44 newton's, block b is 22 newtons, respectively.

00:07 We're asked to determine the minimum weight of c, to see that i've written a block c up there, to keep a from sliding if the coefficient of static friction between a and table is 0 .02, or 0 .20.

00:54 Okay, let's see.

00:59 So prevent sliding, our fs has to equal our t.

01:06 We can block us get t.

01:08 So fs will equal to mu times the force of n, where our force of n will equal wa plus wb, wc, my bad.

01:29 And let's go up.

01:31 Block b.

01:43 For block b, got a way to b.

01:46 Minus our tension equals 0, so t equals wb.

01:55 And t equals our force of static friction, which equals this times this equals coefficient of static friction times wa plus wc.

02:09 So wb equals, and solving for wc, we get wb over the coefficient to static friction minus w .a has to equal 66 newton's.

02:37 So let's go ahead and put our values in.

02:39 How soon i got that.

02:40 So wc equals wb, which was 22, over the co -efficient of static friction, 0 .2 minus 44, and that'll be 2 .2 divided by 0 .2, enter, minus 44, whoops, 22 divided by 0 .2, enter, minus 44 and i get 66.

03:17 Okay, so wc has to equal 66 newton's.

03:24 That's our answer for a.

03:28 Then our question b says block c is suddenly lifted off.

03:44 Block a.

03:47 Find the acceleration of block a if the kinetic friction between a and the table is 0 .15...

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