Question

In Fig. 5-64, a forceof magnitudeis applied to a FedEx box of mass $m_2=1.0\mathrm{kg}$. The force is directed up a plane tilted by $\theta ={37}^o$.The box is connected by a cord to a UPS box of mass $m_1=3.0\mathrm{kg}$on the floor. The floor, plane, and pulley are frictionless, and the masses of the pulley and cord are negligible. What is the tension in the cord?

Short answer

Tension (T) is 4.6 N

Step-By-Step-Solution

Step 2: Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagramdifferent forces acting on the objects can be found. Further, using Newton's second law of motion, the acceleration and tension in the cord can be computed.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\sum Ma$

Step 3: Determining the tension in the cord

Free Body Diagram:

$F-T-m_{2}gs\mathrm{in}\left(\theta \right)=m_2^{}a$

$T=m_{1}a$

By solving above two equations,

$a=\frac{F-m_{2}g\sin \left(\theta \right)}{\left(m+m_2\right)}$

$\begin{array}{l}a=\frac{12-(1\times 98\times sin({37}^a)}{(3+1)}\\ =153{\mathrm{mm}}^2\end{array}$

Now, tension is given by,

$\begin{array}{l}T=m_{1}a\\ =3\times 1.53\\ =4.6N\end{array}$

Hence, tension is 4.6 N

Step-by-step solution

Given information

It is given that,

$F=12N$

$m_2=1.0\mathrm{kg}$

$\theta =3T^o$

$m_1=3.0\mathrm{kg}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagramdifferent forces acting on the objects can be found. Further, using Newton's second law of motion, the acceleration and tension in the cord can be computed.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\sum Ma$

Determining the tension in the cord

Free Body Diagram:

$F-I-m_{2}gs\mathrm{in}\left(\theta \right)=m_{2}a$

$T=m_{1}a$

By solving above two equations,

$a=\frac{F-m_{2}g\sin \left(\theta \right)}{\left(m_1+m_2\right)}$

$\begin{array}{l}a=\frac{12-(1\times 98\times sin({37}^0)}{(3+1)}\\ =1.53m/s^2\end{array}$

Now, tension is given by,

$\begin{array}{l}T=m_{1}a\\ =3\times 1.53\\ =4.6N\end{array}$

Hence, tension is 4.6N