Question

Joe ponder, once used his teeth to lift a pumpkin with a mass of 275 kg. Suppose ponder has a mass of 75 kg, and he stands with each foot on a platform and lifts the pumpkin with an attached rope. If he holds the pumpkin above the ground between the platforms, what is the force exerted on his feet?

Answer 1

The given problem can be exemplified in the following diagram:

Where:

$\begin{gathered} W_j=\text{ Joe's weight} \\ W_p=\text{ weight of the pumpkin} \\ F_f=\text{ force on the fe}et \end{gathered}$

To determine the value of the force on the feed we will use a sum of forces in the vertical direction:

$\Sigma F_y=-W_j-W_p+F_f$

Since there is no movement in the vertical direction the sum of the forces must be equal to zero, therefore:

$-W_j-W_p+F_f=0$

Now we solve for the force on the feet by adding the weight of joe and the weight of the pumpkin in both sides:

$F_f=W_j+W_p$

Now, the weight is the product of the mass by the acceleration of gravity, therefore, we have:

$F_f=m_jg+m_pg$

Now we replace the given values:

$F_f=(75\operatorname{kg})(9.806\frac{m}{s^2})+(275\operatorname{kg})(9.806\frac{m}{s^2})$

Solving the operations we get:

$F_f=3432.10N$

Therefore, the force on the feet is 3432.10 Newtons.