In earlier days, horses pulled barges down canals in the...

In earlier days, horses pulled barges down canals in the manner shown in Fig. 6-63. Suppose the horse pulls on the rope with a force of $7900 \mathrm{~N}$ at an angle of $18^{\circ}$ to the direction of motion of the barge, which is headed straight along the canal. The mass of the barge is $9500 \mathrm{~kg}$, and its acceleration is $0.12 \mathrm{~m} / \mathrm{s}^{2}$. What are the (a) magnitude and (b) direction of the force on the barge from the water?

Key Concepts

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. This fundamental principle connects the motion of an object with the forces acting upon it, allowing us to determine unknown forces when mass and acceleration are known.

Vector Decomposition

Vector decomposition involves breaking a force into its components along mutually perpendicular directions, typically using trigonometric functions such as sine and cosine. This concept is essential when dealing with forces applied at an angle, as it allows one to analyze the effects of each component separately on the motion of an object.

Free-Body Diagrams and Net Force Calculation

A free-body diagram is a graphical illustration that shows all the forces acting on a single object. It serves as a critical tool for setting up equations to compute the net force, which is the vector sum of all individual forces. This method enables the systematic determination of unknown forces, such as resistance or reaction forces from the environment.

Transcript

00:01 Here we have our barge being pulled along by our horse up here with a force of 7 ,900 newtons at an angle of 18 degrees with respect to the slope, or with the river.

00:20 So there's a force from the water, which is going to have to be opposite this force to keep it moving in the x direction.

00:31 So we're told that it's moving in the x direction.

00:35 It has a acceleration of 0 .18 meters per second squared in this x direction.

00:46 So this is x and then y is up.

00:50 So we need to figure out what that force of the water is and the direction that it is in.

00:55 So we're going to split this into two parts.

00:58 We're going to have the x forces and the y forces.

01:01 Let's start with the y forces.

01:05 The barge is not accelerating in the y direction, so the net force is going to be zero.

01:11 And we're going to have the y component of the applied force.

01:16 So that is f times the sign of that angle minus the y component of the water's force.

01:28 So i'm going to let's call that fwy.

01:32 So this is the force that the water is applying from the river.

01:40 So these two forces have to be equal and opposites.

01:43 It's equal, since they combine to equal zero.

01:47 So let's just say that.

01:49 We're just adding this to both sides.

01:53 Now, in the x direction, we've got a little bit more going on.

01:58 Here we do have an acceleration, so the net force is not zero.

02:03 And we have the x component of this force, so that is force times the cosine of that angle, and that's going to be minus the force from the water in the x direction, fwx...

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