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Two friends are carrying a 200-kg crate up a flight of stairs. The crate is 1.25 m long and 0.500 m high, and its center of gravity is at its center. The stairs make a 45.0° angle with respect to the floor. The crate also is carried at a 45.0° angle, so that its bottom side is parallel to the slope of the stairs (Fig. 2). If the force each person applies is vertical, what is the magnitude of each of these forces?

Sagot :

Ajeigbeibraheemidn

Let consider the whole system to be at equilibrium where the net force is zero. The forces [tex]F_A[/tex] and [tex]F_B[/tex] are acting in the upward direction while the weight (w) of the crate is acting in the downward direction.

The force at equilibrium in the y-direction is computed as:

[tex]\sum F_y = ma_y[/tex]

[tex]F_A +F_B= mg[/tex]

[tex]\mathbf{F_A = mg -F_B}[/tex]  ---- (1)

Recall that the weight w = mg

where;

  • mass (m) = 200 g
  • g(constant gravity) = 9.8 m/s²

w = 200 × 9.8

w = 1960 N

The torque about the bottom end is also zero. i.e.

[tex]\sum \tau=0[/tex]

∴

[tex]F_B L_Z - wL_w = 0[/tex] --- (2)

[tex]F_B L_Z= wL_w[/tex]

[tex]F_B = \omega (\dfrac{L_w}{L_z})[/tex]

[tex]F_B =1960\times (\dfrac{0.375\times cos 45^0}{1.25\times cos 45^0})[/tex]

[tex]F_B =1960\times (0.3)[/tex]

[tex]\mathsf{F_B = 588 \ N}[/tex]

Replacing the value for the force of B into equation (1); we have:

[tex]F_A = mg - F_B[/tex]

[tex]F_A = 1960 - 588[/tex]

[tex]\mathsf{F_A = 1372 N}[/tex]

Thus, we can conclude that the magnitude of these forces i.e force B and force A are 588 N and 1372 N respectively.

Learn more about the equilibrium of forces here:

https://brainly.com/question/24018969?referrer=searchResults

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