Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center?

Question

Diagram showing forces F1, F2, and F3 acting on a wheel for torque analysis.

Accepted Answer

Hi everyone. Today, we are going to determine the resultant torque from the three forces acting upon a disc whose diameter is 1.5 m, or whose radius is actually 0.75 m right here, Which actually has three different Children pulling on it with three different angles and three different forces. So the first child is going to be pulling out of 49° relative to the radius, the second pulling parallel to the radius, and the last one pulling perpendicular to it. And we want to determine the resultant torque resulting from the three forces about an axle through the center of the disk and perpendicular to it. So remember that the result on torque formula is going to be represented by this sigma tau right here, which is going to be tell of everything going counterclockwise minus style of everything going clockwise. So recall the last time we determine the convention of everything going counterclockwise to be a positive torque while everything going clockwise to be a negative torque. Now we can actually recall that the torque formula itself is going to be calculated by multiplying the distance represented by our right here with the force exerted upon it well represented by the f right here, but I remember that this to have to actually be perpendicular to one another. So therefore if not we can add this angle correction here, which is going to be sine of theta. Now we can actually just jump right into this diagram right here and start analyzing each force first. We want to start with the first child. So talk number one right here. We know that force one here is not actually perpendicular to the radius here. So what we want to do is to actually do a projection of force one. So this component right here is actually just going to be F one multiplied by sine of 49 degrees. Or sign off the angle opposite to the projected or projected component itself. So now we can actually use this and put it into the torque formula which is just going to be our multiplied by F one times sine of 49 degrees or sign of data equals two B the R which is 0.75 m times 10.4. Newton of the fourth one which is known in this example diagram here multiplied by sine of 49. This will actually be 5.89. Newton Times meter. Next we want to look on to force to right here or child to which in this case child to is actually pulling perpendicular. Oh no parallel parallel to the radius itself like. So, so because of that we actually know that the torque formula itself can only be calculated when the radius and the force are actually perpendicular to one another and therefore when it is peril then we can determine that the torque will be zero. However, we can just plug them into the torque formula that we know to actually prove that it is going to be zero. So in this case we want to use the sign or the correction angle of a sign zero because remember that it is um pulled parallel to the radius itself and sign zero is going to be zero. So tao two is going to be zero. Newtons times meter. Now, for the last shot right here, We can see that F3 is actually already parallel perpendicular to the radius so because F Yeah, F three is already perpendicular to the radius itself, we can directly put them in, we can neglect the correction angle right here. But if you want to write it down for consistency purposes, we can put side of 90 degrees because it is already perpendicular, which is sign of 90 is just 01 right here, It's just one. So now we can put in these values which is 0.75 m for the R&26. Newton for the force multiplied by one. And this will actually come out to be 19.5 newtons times meter. Now we have to one tau two and three, but we want to indicate which tower or which torque is actually going counterclockwise and which torque is actually going clockwise. We know that F one right here is pulling to pulling the disk to this direction right here, which is a direction counterclockwise, which is positive. Well, we know that three is actually going right pulling this way, which is a direction clockwise, which is going to be negative. So now we want to combine them into our results and torque formula right here, which is just going to be talk of everything going counterclockwise, when the start of everything going clockwise, which is going to be counterclockwise, Park one Minour Clockwise, Top three. And we just plucking plucking these two values right here, 5.89 -19.5, which is going to be minus 13. newtons times meter and this minus 13.6 newton status meter right here is actually going to be our answer, which is actually answer option B and there you go.

Verified Solution

Key Concepts

Torque

Net Torque

Force Components