What is the force required to tilt a rectangular object when exerted at 2 different locations?

Question

I have a question to confirm if my "hunch" is correct. Assuming 2 scenarios of similar force but is exerted at different height of a rectangular object with a pivot point (shown in the pic below). Based on Torque (perpendicular force * distance from the pivot point), it should be easier to tilt the object from F1 location vs F2 location assuming a similar amount of Torque needed to lift the object. Torque=d1 * F1 = d2 * F2 ). Based on that formula, then F1<F2 to get same torque.

However, when I use the 2nd class lever equation, I find F1 ==F2 based on the equation below: Assuming the CoM is in the middle of the object, using the 2nd class lever formula, i.e. d1 * F1= mg(CoM)* 0.5 *d1) , and so F1 = (mg(C0M) * 0.5 * d1)/ d1. I used the same formula for F2 and I found out F1=F2. Is this the exact force to tilt the object and if so, why both F1=F2 when it should be F1 < F2, or am I confusing force and torque?

Answer 1

First of all, torque is the product of force and the distance of a point from the pivoting point perpendicular to the direction force.So, $${\tau} = {\vec r}\;×\;{\vec F}$$ $${\tau} = |{\vec r}||{\vec F}|sin{\theta}$$ where ${\theta}$ is the angle between $\vec F$ and $\vec r$.

So, you must first resolve the forces.

Now, for the line of length ${d_1}$ ,

Angle between $\vec {F_1}$ and $\vec r$ is $\alpha$

Component of $\vec {F_1}$ perpendicular to $\vec r $(along $d_1$)= $${F_1}sin{\alpha}$$ Angle between ${mg}$ and $\vec r$ is $$(90°-\alpha)$$ Component of $mg$ perpendicular to $\vec r$= $$mgcos{\alpha}$$ So, net torque: $${\tau}= {F_1}{d_1}{sin\alpha}-{m}{g}{cos\alpha}(\frac{d_1}{2})$$

Similarly for the other line of force, $${\tau}= {F_2}{d_2}{sin\beta}-{m}{g}{cos\beta}(\frac{d_2}{2})$$

Now talking of $2^{nd}$ class lever, The formula that you have used,i.e., $${Effort}×{Effort\;arm}={Load}×{Load\;arm}$$ Only applies for balanced situation (rotational equilibrium).In that case, $${F_1}{sin\alpha}({d_1})={mg}{cos\alpha}(\frac{d_1}{2})$$ $${F_2}{sin\beta}({d_2})={mg}{cos\beta}(\frac{d_2}{2})$$

But, as you are lifting of the block and rotating it around a pivot point (say anticlockwise), you cannot apply that equation for second class levers as the stick is not in rotational equilibrium. So, in this case(for anticlockwise rotation): $${F_1}{sin\alpha} > \frac{{mg}{cos\alpha}}{2}$$ (I have cancelled $d_1$ from both sides) $${F_2}{sin\beta} > \frac{{mg}{cos\beta}}{2}$$ (cancelled $d_2$ from both sides)

$\alpha$ and $\beta$ are going to vary constantly as you lift the block. So, you would need exact values to calculate the torque at any instant. But you are going to get the same relation between $F_1$ and $F_2$ at any instant. You need not worry about the second class lever equation as it is just an extrapolation of the balanced torque condition. I hope you have got the idea of how to frame the formulas.

Answer 2

I see you have got the definition of torque wrong. It is not just force multiplied by distance from pivot point. Torque is distance from pivot point multiplied by the amount of force which is perpendicular to the direction of point of application from pivot point. It can also be stated as the full force multiplied by the perpendicular distance . So in our case h1f1= mglength of block/2=h2*f2 and we cannse f1 is not equal to f2 and ever which is bigger. You can also try taking cross products of displacement vector and force vector.