Block A in Fig. P5.89 has mass 4.00 kg, and block B has mass...

Block $A$ in $\textbf{Fig. P5.89}$ has mass 4.00 kg, and block $B$ has mass 12.0 kg. The coefficient of kinetic friction between block $B$ and the horizontal surface is 0.25. (a) What is the mass of block $C$ if block $B$ is moving to the right and speeding up with an acceleration of 2.00 $m/s^2$? (b) What is the tension in each cord when block $B$ has this acceleration?

Key Concepts

Tension in Cords

Tension is the force transmitted along a string, rope, or cord when it is pulled tight by forces acting from opposite ends. In an ideal system with massless cords, the tension is consistent along each individual section of the cord, and determining the tension in each segment is crucial for solving the dynamics of the connected system.

Systems of Connected Bodies

When dealing with objects connected by cords or ropes, it is important to consider how the motion of one body affects the others. This concept involves writing separate equations for each object and then combining them, taking into account constraints such as the uniform acceleration of connected objects.

Kinetic Friction

Kinetic friction is the resistive force that acts opposite to the direction of motion between two sliding surfaces. It is quantified by the coefficient of kinetic friction multiplied by the normal force, and it plays a critical role in determining the net force on a moving object.

Newton's Second Law

This is the fundamental principle that states the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In problems involving connected bodies, it is used to set up equations for each body and solve for unknown quantities such as mass, acceleration, or tension.

Free-Body Diagrams

A free-body diagram is a visual representation that isolates an object and shows all the forces acting upon it. This tool is essential for correctly applying Newton's laws, especially when analyzing systems with multiple forces and connected bodies.

Transcript

00:01 In this problem, we're going to do parts a and b together.

00:03 So first, i'm going to start with block a and draw the free body diagram.

00:08 We have a tension force, tension between a and b, pulling it up, and we have massive a times gravity pulling it down.

00:17 We also have an acceleration pulling it up.

00:21 And so if i apply newton's second law in the y direction, i can't say it equal to zero because we have an acceleration.

00:30 So this full equation implies that tension between a and b, minus mass of a times g is equal to mass of a times a.

00:40 Solving this for the tension between a and v, i get mass of a times a plus g.

00:50 We know what g is a is given, and mass of a is also given.

00:54 And so this is four times two plus 9 .8, which is equal to 47 .2 nons.

01:04 Now i'm going to do a free body diagram for b.

01:09 And so if this is block b, i have a tension between b and c, pulling it to the right.

01:16 I have a normal force upward.

01:18 I have two forces to the left.

01:21 One is tension between a and b, and the other is friction force, kinetic friction force, which is equal to that.

01:33 And then i also have downward gravity, which is equal to the mass of b times g.

01:41 So the sum of the forces here in the y direction does equal zero because there's no acceleration in the y direction.

01:48 This implies that n minus mbg is equal to zero.

01:53 So n is equal to mbg.

01:58 Now i can use this expression here to solve for the friction force.

02:02 So that means the friction force is equal to mu -sup k, which is given to us at 0 .25 times n, but n is mb, which is 12, and then g, which is 9 .8...

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