Question

Matt has a portable basketball goal in his driveway. He has the basket set 8 ft (2.4 m) high so he can practice dunking the ball. He slams the ball through the hoop and then hangs onto the rim. This exerts a downward force of 600 N on the front of the rim. The front of the rim is 1.1 m in front of the front edge of the portable basketball goal%u2019s base. The mass of the whole portable basketball goal is 70 kg. The center of gravity of the portable basketball goal is 1.0 m behind the front edge of its base.

a. How much torque is produced around the front of the goal base by the 600 N force Matt exerts on the front of the rim?
b. How much torque would be needed to tip the goal?
c. What is the largest vertical force that can be exerted on the front edge of the rim before the portable basketball goal begins to tip?

Answer 1

a) τ₁ = 660 N m, b) τ’= 686 N m, c) F = 623.6 N

Step-by-step explanation:

a) For this exercise let's use the concepts of torque and rotational balance.

For this we set a reference system at the base and assuming that the counterclockwise rotations are positive

where the force F = 600 N, the distance to the axis is x = 1.1 m, the mass of the system m = 70g and the weight is placed at the point of the center of gravity x_{cm} = -1.0 m

The torque at the front is

τ₁ = F x

τ₁ = 600 1.1

τ₁ = 660 N m

b) let's write the rotational equilibrium condition

∑ τ = 0

τ'- W x_{cm} = 0

τ ’= mg x_{cm}

τ’= 70 9.8 1.0

τ’= 686 N m

c) the greatest force Matt can apply

τ’= F x

F = τ’/ x

F = 686 / 1.1

F = 623.6 N