Question

A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press

Answer 1

Answer:

$\mathbf{F_1=4.41*10^4\ N}$

$\mathbf{F_2 = 1.176*10^5 \ N}$

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

$\sum F = 0$ and $\sum \tau = 0$

From the image, expressing the forces through the y-axis, we have:

$F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies \mathtt{1.617\times 10^5 \ N}$

Also, let the force $F_1$ be the pivot and computing the torque to determine $F_2$:

Then:

$F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P$

$F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}$

$F_2 = 117600 \ N$

$\mathbf{F_2 = 1.176*10^5 \ N}$

For the force equation:

$F_1+F_2=1.617*10^5 \ N;$

where:

$F_2 = 1.176*10^5 \ N$

Then:

$F_1+1.176*10^5 \ N=1.617*10^5 \ N$

$F_1=1.617*10^5 \ N-1.176*10^5 \ N$

$F_1=44100\ N$

$\mathbf{F_1=4.41*10^4\ N}$