# High school Momentum/Torque problem [closed] This is embarrassing but I can't seem to solve this problem, even though I know how to resolve forces involved.

The weight of the base is 868 N.

The problem should also state "assume the assume the point where the base remains on the floor does not move". Consider the torque to turn the set over, about the fixed point of the base on the floor.

Using the 15 degree angle in the figure, you can determine the angle of the base relative to the floor, the angle of the weight relative to the inclined post, and the angle of the wind force relative to the backboard. Note the angle between the base and post is 90 degrees. Vary the given 15 degrees from 0 to 90 degrees to help see how the other three angles change and are related to this 15 degree angle.

The points where the forces act determine moment the arms (distance from the fixed point of the base on the floor to a force) of the forces for torque. You need the sin of the angle between the moment arm and the force to determine the torque of each of the three forces.

For example. the torque from the weight of the base is $$\vec r \times \vec F_{base}$$ which is (0.45)(868) sin(90 -15) clockwise. Similarly figure out the other torques; when total counter-clockwisewise = total clockwise torque the set starts to rotate over.

There is a force acting where the base is fixed on the floor, but it does not contribute to the torque about that point, because of the zero moment arm for this force.

Note torque (and angular momentum) depend on the point about which they are evaluated. Here we use the point where the base stays on the floor so we do not have to evaluate the torque from the force at that point.

• I think I'm confused about the angles, there are 3 points and I can't seem to figure out what to do. Jun 20 at 18:13
• Yeah the force of the wind is 158 N. Jun 20 at 18:49
• Hey you could also edit the answer and show the steps so that I can accept your answer. This really is the hardest problem I've done. Jun 20 at 18:54
• I did so. Note torque (and angular momentum) depend on the point about which they are evaluated. Here we use the point where the base stays on the floor so we do not have to evaluate the torque from the force at that point. Jun 20 at 19:25

A torque $$T$$ is given as:

$$T = F x_{\perp}$$

where $$F$$ is the force and $$x_{\perp}$$ is the perpendicular distance from the acting force to the pivot of it's axis.

The force of the wind will start to topple the basket ball set (push it over) when the moments of force balance each other. Summing the moments of force or torques in the clockwise and counter-clockwise directions, gives:

$$\sum T_{clockwise} = \sum T_{counter \ clockwise}$$