Question

1. a 15 kg object rests on a table. a cord is attached to this object and also to a wall. another object is hung from this cord as shown. if the coefficient of friction between the 15 kg object and the table is 0.27, what is the maximum mass that can be hung, without any movement?

Explanation:

Step1: Calculate the normal force on the 15 - kg object

The normal force $N$ on the 15 - kg object on the table is equal to its weight since there is no vertical acceleration in the y - direction for the 15 - kg object. Using the formula $N = mg$, where $m = 15$ kg and $g=9.8$ m/s², we have $N = 15\times9.8=147$ N.

Step2: Calculate the maximum static friction force

The maximum static friction force $f_s$ is given by the formula $f_s=\mu_sN$, where $\mu_s = 0.27$ and $N = 147$ N. So $f_s=0.27\times147 = 39.69$ N.

Step3: Analyze the forces on the hanging object

Let the mass of the hanging object be $m$. The tension $T$ in the cord has a horizontal and a vertical component. At the maximum - mass condition, the horizontal component of the tension $T_x$ is equal to the maximum static friction force $f_s$. The vertical component of the tension $T_y$ is equal to the weight of the hanging object $mg$. If the angle between the cord and the horizontal is $\theta = 30^{\circ}$, then $T_x=T\cos\theta$ and $T_y = T\sin\theta$. Since $T_x=f_s$, we have $T=\frac{f_s}{\cos\theta}$. And since $T_y = mg$, we have $mg=T\sin\theta$. Substituting $T=\frac{f_s}{\cos\theta}$ into $mg=T\sin\theta$, we get $mg=\frac{f_s}{\cos\theta}\sin\theta=f_s\tan\theta$.

Step4: Solve for the mass of the hanging object

We know $f_s = 39.69$ N and $\theta = 30^{\circ}$, so $m=\frac{f_s\tan\theta}{g}$. Substituting the values, $\tan30^{\circ}=\frac{\sqrt{3}}{3}\approx0.577$ and $g = 9.8$ m/s². Then $m=\frac{39.69\times0.577}{9.8}=\frac{22.90}{9.8}\approx2.34$ kg.

Answer:

$2.34$ kg