Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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Question:

Two objects with masses of 3.00 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley, as in Figure P4.66. Determine (a) the tension in the string, (b) the acceleration of each object, and (c) the distance each object will move in the first second of motion if both objects start from rest.

Figure P4.66

Explanation

Explanation:

Given info: The masses are m1$m_1$= 3.00 kg and m2$m_2$= 5.00 kg.

The free body diagram is given below.

From the free body diagram,

T−m1g=m1a$T-m_{1}g=m_{1}a$ (I)

m2g−T=m2a$m_{2}g-T=m_{2}a$ (II)

• m1$m_1$ and m2$m_2$ are the masses of blocks.
• a is the acceleration.
• g is the acceleration due to gravity.
• T is the tension in the cable.

Re-arrange Equation (II) to get a.

a=g−(Tm2)$a=g-\left(\frac{T}{m_2}\right)$ (III)

Substitute Equation (III) in (I) and re-arrange to get T.

T=g(2m1m2m1+m2)$T=g\left(\frac{2m_1{m}_2}{m_1+m_2}\right)$

Substitute 9.8 ms−2$9.8{\text{ms}}^{-2}$ for g, 3.00 kg for m1$m_1$ and 5.00 kg for m2$m_2$ to get T.

T=(9.8 ms−2)(2(3.00 kg)(5.00 kg)(3.00 kg)+(5.00 kg))=36.75 N≈36.8 N$\begin{array}{c}T=\left(9.8{\text{ms}}^{-2}\right)\left(\frac{2\left(3.00\text{kg}\right)\left(5.00\text{kg}\right)}{\left(3.00\text{kg}\right)+\left(5.00\text{kg}\right)}\right)\\ =36.75\text{N}\\ \approx 36.8\text{N}\end{array}$

Conclusion:

The tension in the string is 36.8 N.