Chapter 4: Problem 68

In a playground, two slides have different angles of incline $\theta_{1}$ and $\theta_{2}\left(\theta_{2}>\theta_{1}\right) .$ A child slides down the first at constant speed; on the second, his acceleration down the slide is $a .$ Assume the coefficient of kinetic friction is the same for both slides. (a) Find $a$ in terms of $\theta_{1}, \theta_{2},$ and $g.$ (b) Find the numerical value of $a$ for $\theta_{1}=45^{\circ}$ and $\theta_{2}=61^{\circ}.$

Short Answer

Expert verified

Answer: The acceleration of the child down the second slide is $$a=g(\frac{\sqrt{3}-1}{2})$$, where g is the gravitational constant and the angles$$ $\theta_{1}=45^{\circ}$ and $\theta_{2}=61^{\circ}$.

Step by step solution

Set up friction equations for both slides

We need to find the frictional force for both slides. We know that the frictional force (F_f) equals the normal force (F_N) multiplied by the coefficient of friction. We also know that the normal force equals the component of the gravitational force that acts perpendicular to the slide. So, we get the following relationships for both slides: $$F_{f_{1}} = \mu F_{N_{1}} = \mu mg\cos{\theta_{1}}$$ $$F_{f_{2}} = \mu F_{N_{2}} = \mu mg\cos{\theta_{2}}$$

Set up Newton's Second Law equations for both slides

For slide 1, since the child's speed is constant, the net force is zero. For slide 2, the net force acting on the child is responsible for the acceleration a. This gives us the following equations: Slide 1: $$F_{g1} - F_{f_{1}} = 0$$ $$mg \sin{\theta_{1}} - \mu mg\cos{\theta_{1}}= 0$$ Slide 2: $$F_{g2} - F_{f_{2}} = ma$$ $$mg \sin{\theta_{2}} - \mu mg\cos{\theta_{2}}=ma$$

Solve for the acceleration a

We have two equations and two unknown variables: a and μ. First, we solve for μ using the Slide 1 equation: $$\mu = \frac{\sin{\theta_{1}}}{\cos{\theta_{1}}}$$ Now, we substitute this expression for μ into the Slide 2 equation and solve for a: $$mg\sin{\theta_{2}} - \frac{\sin{\theta_{1}}}{\cos{\theta_{1}}} mg\cos{\theta_{2}}=ma$$ $$a=g(\sin{\theta_{2}}-\frac{\sin{\theta_{1}}\cos{\theta_{2}}}{\cos{\theta_{1}}})$$

Use given angles to find the numerical value of a

Now that we have an expression for a, we can find the numerical value using the given angles: $$a=g(\sin{61^{\circ}}-\frac{\sin{45^{\circ}}\cos{61^{\circ}}}{\cos{45^{\circ}}})$$ First, we find the trigonometric values for the given angles: $$\sin{45^{\circ}} = \sin{61^{\circ}} = \frac{\sqrt{2}}{2}$$ $$\cos{45^{\circ}} = \cos{61^{\circ}} = \frac{\sqrt{3}}{2}$$ Next, we plug these values into our equation for a and simplify: $$a=g(\frac{\sqrt{3}}{2}-\frac{\frac{\sqrt{2}}{2}\frac{1}{2}}{\frac{\sqrt{2}}{2}})$$ $$a=g(\frac{\sqrt{3}}{2}-\frac{1}{2})$$ Therefore, the acceleration of the child down the second slide is: $$a=g(\frac{\sqrt{3}-1}{2})$$

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Short Answer

Step by step solution

Set up friction equations for both slides

Set up Newton's Second Law equations for both slides

Solve for the acceleration a

Use given angles to find the numerical value of a

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