A 2.8 meter long board sits across two sawhorses as shown in the picture to the

Question

A 2.8 meter long board sits across two sawhorses as shown in the picture to the right.) The left end of the board hangs off of the edge by a distance of 0.6 meters. The board has a mass of 3.5 kg. What is the weight of the board? If the board is of unioform density, where is its center of gravity of gravity located? Draw a free-body diagram that identifies all of the forces on the board and their locations. What is the net force acting on the board? What is the net torque acting on the board? What is ihe normal force caused by the sawhorse on the left? What is the normal force caused by the sawhorse on the right? An adorable toddler crawls to the left end of the board. How much mass can the toddler have without the board tipping over?

Explanation / Answer

Hi,

I suppose that you want to answer the three last questions. If that's the case then we have the following:

In order to find the normal forces exerted by the sawhorses at each side we use the fact that the net force and the net torque over the boar are equal to zero. For convenience we choose the axis of the torque in a way that crosses just in the center of the board (the center of gravity), so the torque due to the weight is cero as well.

Balance of forces.    W = N1 + N2

Balance of torques.   N1(0.8 m) = N2(1.4 m) :::::: N1 = 1.75 N2

W = 2.75 N2 ::::: N2 = W/2.75 = (mg)/2.75 = (3.5 kg)(9.8 m/s2)/ 2.75 = 12.47 N ; this is the normal force to the right.

N1 = mg - N2 = (9.8 m/s2)(3.5 kg) - 12.47 N = 21.83 N; this would be the normal force to the left.

Note: it is expected that N1 (to the left) is bigger than N2 (to the left) because the first one has to produce the same torque as N2 while having a smaller distance to the axis.

If we now have a toddler at the end of the board, we have a new force that points in the same direction than the weight of the board. Therefore we have to make new balances of torques and forces; although, in this case, it would be better to fix the axis over the sawhorse to the left.

Balance of forces.   F + W = N1 + N2

Balance of torques. (0.6m)F + (1.4m)N2 = (0.8m)W

It is obvious that there is a different number of unknowns and equations. If we consider all the previous results, then we have only an unknown (F) and two equations; but, if we consider that the only value we can preserve is the weight of the board, then we have three unknowns (F, N1, N2) and two equations.

However, we can consider that the moment when the board is just about to tipping over N2 is cero beacuse the board will not be touching the sawhorse to the right.

Balance of forces.   F + W = N1 + N2 :::::::::::::::::::::::::::::::::::: F + W = N1

Balance of torques. (0.6m)F + (1.4m)N2 = (0.8m)W ::::::::::::::: (0.6m)F = (0.8)W :::::::. F = 1.33 W

F = 1.33 (9.8 m/s2)(3.5 kg) = 45.62 N

2.33 W = N1 ::::::: N1 = (2.33) (9.8 m/s2) (3.5 kg) = 79.92 N

The mass of the toddler can be found using the gravity and the weight found (F):

m = F / g = (45.62 N) / (9.8 m/s2) = 4.66 kg

I hope it helps.

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