Physics A heavy object of mass m is to be dragged along a horizontal surface by a rope making an

step 1 Here we're given an equation for the force. So f is equal to C times m times g over C sine theta

Physics A heavy object of mass m is to be dragged along a...

Key Concepts

Optimization in Physics using Calculus

Many physics problems require finding the minimum or maximum of a quantity, such as the least force needed to achieve motion. This is typically approached by expressing the quantity as a function of a variable, then using calculus—taking the derivative and setting it to zero—to find critical points. By analyzing these points, one can determine the conditions under which the required force is minimized or maximized.

Force Decomposition

In problems involving the dragging of objects, forces are often decomposed into their horizontal and vertical components. This decomposition is crucial because the vertical component of the applied force can alter the normal force, which in turn affects the frictional force opposing the motion. Analyzing the problem via vector components allows one to correctly set up equations for both the net horizontal force and the equilibrium of the vertical forces.

Friction and Normal Force

Friction is a resistive force that depends on the normal force and the coefficient of friction. When an applied force has a vertical component, it reduces the normal force (which is typically the weight of the object) and consequently lowers the frictional force. The coefficient of friction is a material-dependent constant that, when multiplied by the normal force, gives the magnitude of the frictional force. This interplay is fundamental in understanding how the angle of application affects the force required to move the object.

Trigonometric Relationships in Force Analysis

Trigonometric functions are used to relate the components of forces when the force is applied at an angle. In this context, sine and cosine functions help resolve the force into its vertical and horizontal components, respectively. These relationships are critical when establishing the equations that govern the balance of forces, and they often lead to conditions like the one derived (tan ? = c) for optimal configurations.

Transcript

00:01 Here we're given an equation for the force.

00:03 So f is equal to c times m times g over c sine theta plus cosine beta.

00:18 Okay.

00:19 So the first thing that you need to know is that c, m, and g are all constants.

00:26 So that means our only variable here is the angle theta.

00:29 So which means f is actually just a function of theta.

00:32 Okay, so if we're trying to find the minimum force required, then we're, then we want to find the critical values, right? so which means we want to take the derivative, and we're taking the derivative of f with respect to theta.

00:48 Okay, so meaning we're looking for f prime of theta.

00:56 Okay, so then this is equal to, so this is just c times m times g because those are all constants.

01:03 And then we have this thing on the bottom, which we can say that it's really the power of negative 1.

01:09 Right.

01:09 So this is really, we can pull the negative 1 to the front.

01:13 And then we have c, sine theta plus cos theta to the power of now negative 2.

01:19 And then chain rule.

01:20 So we need to multiply by the derivative of the inside.

01:24 So that's going to be c times cost theta.

01:28 And then derivative of course theta is negative sine theta.

01:35 Okay.

01:36 So now let's rewrite this.

01:38 This is equal to.

01:41 On the top, we have negative.

01:45 Let's cancel that out.

01:46 So this is c .m .g.

01:48 We'll reverse that.

01:48 So this is sine theta minus c times cost theta.

01:55 And this is over c sine theta plus cosine theta.

02:06 So now we can see that, well, this thing on the bottom, since we have it being squared, this is going to be always, non -negative.

02:15 So we can see that it doesn't exist if this thing is equal to zero.

02:22 Okay, so we have f prime of theta does not exist when c -sign theta plus cost theta is equal to zero.

02:36 Okay, so this is one of our critical values.

02:39 And the other critical value would be f prime of theta is equal to zero when our top is equal to zero.

02:47 All right.

02:48 So since these are constants, we just get rid of them.

02:50 So that would be when sine theta minus c times close theta is equal to zero.

02:58 Okay, so we have potentially two different critical values.

03:02 So these are the critical values.

03:09 Okay, so let's go ahead and find them.

03:12 So let's say we have, let's do this one first.

03:16 Okay.

03:16 So from here, we have, let's say, yeah, so we'll do it as this.

03:27 So we have cost - theta is equal to negative c -sign theta, dividing through by cost -theta.

03:36 We have 1 is equal to negative -c tangent theta, so which means we have 10 -theta is equal to negative -1 over -c...

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