A suitcase of weight \(M g=450\). N is being pulled by a small strap across a level floor. The coefficient of kinetic friction between the suitcase and the floor is \(\mu_{k}=0.640 .\) a) Find the optimal angle of the strap above the horizontal. (The optimal angle minimizes the force necessary to pull the suitcase at constant speed.) b) Find the minimum tension in the strap needed to pull the suitcase at constant speed.

To understand the forces acting on the suitcase, we need to first draw a free body diagram. The forces acting on the suitcase are: the gravitational force (or weight) \(Mg\) acting downwards, the normal force \(N\) exerted by the floor acting upwards, the tension force \(T\) exerted by the strap, and the friction force \(f\) acting opposite to the direction of motion.

We can represent the tension force acting on the suitcase as two components: - The horizontal component, denoted by \(T_x = T\cos{\theta}\), where \(\theta\) is the angle between the strap and the horizontal. - The vertical component, denoted by \(T_y = T\sin{\theta}\).

For the suitcase to be in equilibrium in the vertical direction, the forces should cancel each other out. Thus, the normal force \(N\) is equal to the sum of the vertical component of the tension force and the gravitational force: $$N = T_y + Mg$$ $$N = T\sin{\theta} + Mg.$$ In the horizontal direction, the kinetic friction force \(f\) should be equal to the horizontal component of the tension force: $$f = T_x$$ $$f = T\cos{\theta}.$$

We know that the friction force is given by: $$f = \mu_{k}N.$$ From Step 3, we know that \(f = T\cos{\theta}\). Therefore, $$T\cos{\theta} = \mu_{k}N.$$

To find the optimal angle, we'll make use of the formula for the horizontal component of tension that we found in Step 4. $$T\cos{\theta} = \mu_{k}N.$$ Now, divide both sides by \(\cos{\theta}\): $$T = \frac{\mu_{k}N}{\cos{\theta}}.$$ To minimize \(T\), we can take the derivative with respect to \(\theta\) and set it to zero: $$\frac{d}{d\theta}\left(\frac{\mu_{k}N}{\cos{\theta}}\right) = 0.$$ Using quotient rule, the above expression can be written as: $$\frac{-\mu_{k}N\sin{\theta}}{\cos^2{\theta}}=0.$$ The optimal angle is attained when \(\sin{\theta}=0\). From trigonometric identities, we know that this occurs when: $$\theta = \tan^{-1}(\mu_{k}).$$ Substitute the given values for \(\mu_{k}=0.640\), $$\theta = \tan^{-1}(0.640) \approx 32.7^\circ.$$ Thus, the optimal angle of the strap above the horizontal is approximately \(32.7^\circ\).

Now that we know the optimal angle \(\theta\), we can use the equation derived in Step 4 to find the minimum tension \(T\): $$T\cos{\theta} = \mu_{k}N.$$ We also have the relation \(N = T\sin{\theta} + Mg\) from Step 3. We can solve these two equations simultaneously to get the value of \(T\). Rearranging the equation for \(N\) and substituting the value of \(f\) found earlier: $$N = \frac{Mg}{\sin{\theta} - \mu_{k}\cos{\theta}}.$$ Now substitute the value of \(N\) into the equation for \(T\): $$T\cos{\theta} = \mu_{k}\frac{Mg}{\sin{\theta} - \mu_{k}\cos{\theta}}.$$ Divide by \(\cos{\theta}\) and use the given values \(Mg = 450\,\text{N}\) and \(\mu_{k}=0.640\) in the equation: $$T = \frac{(0.640)(450)}{\sin{(32.7^\circ)} - (0.640)\cos{(32.7^\circ)}}.$$ After solving the above equation, we get: $$T \approx 233.8\,\text{N}.$$ Therefore, the minimum tension in the strap needed to pull the suitcase at constant speed is approximately \(233.8\,\text{N}\).