Question

The 21 kg mass is attached by a cord to a mass hanging over the edge of the table. The frictional force between the mass and the table's surface is 121 N. The mass accelerates at a rate of 0.73 m/s^2 to the right. What is the weight of the hanging block?

Answer 1

Answer:

147.3 N

Explanation:

Two-Mass Systems

To solve a system where two masses are interacting with each other, we must set up the formulas by applying Newton's second law for each mass. Then, we find the required magnitudes by solving a system of equations.

Our system consists of a hanging object with mass $m_2$ attached to an object of mass $m_1$ lying in a table which applies a known friction force of $F_R=121 N$. We also know the system accelerates at $0.73\ m/sec^2$. The situation is pictured in the image below.

Analyzing the forces acting upon mass $m_1$ we have, in the horizontal axis, where movement is taking place:

$\displaystyle T-F_R=m_1\ a$

Where T is the rope's tension force. Now taking the vertical axis of the second mass, we have

$\displaystyle T-W_2=-m_2\ a$

The acceleration is negative since it's directed downwards, contrary to the positive default direction (right and up). Subtracting both equations:

$\displaystyle W_2-F_R=m_1\ a+m_2\ a$

Solving for $W_2$

$\displaystyle w_2=F_R+m_1\ a+m_2\ a$

We know that

$\displaystyle W_2=m_2\ g$

so, the above formula becomes

$\displaystyle m_2\ g=F_R+m_1\ a+m_2\ a$

Rearranging and factoring

$\displaystyle m_2(g-a)=F_R+m_1\ a$

Solving for $m_2$

$\displaystyle m_2=\frac{F_R+m_1\ a}{g-a}$

Let's use our known data

$\displaystyle m_2=\frac{121+21(0,73)}{9,8-0,73}=\frac{136.33}{9.07}$

$\displaystyle m_2=15.03\ kg$

Finally, we compute the object's weight

$\displaystyle W_2=m_2.g=15.03(9.8)=147.3\ N$