Answer: the object will have an acceleration and the object's velocity will change.
Explanation: I just know
Answer:
a. 1715 N b. 2401 N
Explanation:
Let F = force due to calf muscle, F' = force due to tibia and N = force due to ground = weight of man = mg where m = mass of man = 70 kg and g = acceleration due to gravity = 9.8 m/s².
a. Magnitude of the forces exerted on the foot by the muscle
Since the force due to the calf muscle is 6.0 cm behind the ankle joint and the normal force due to the ground is 15.0 cm in front of the ankle joint and the force due to the tibia is at the ankle joint, taking moments about the ankle joint,
F × 6 cm + F' × 0 cm = N × 15 cm
6F = 15N = 15mg
F = 15mg/6
= 15 × 70 kg × 9.8 m/s²/6
= 1715 N
b. Magnitude of the forces exerted on the foot by the tibia
Taking moments about the calf muscle force, we have
F × 0 cm + F' × 6 cm = N × (15 cm + 6 cm)
6F' = 21N = 21mg
F' = 21mg/6
= 21 × 70 kg × 9.8 m/s²/6
= 2401 N
Answer:
See the attached image and the explanation below
Explanation:
We must draw a schematic of the described problem, after the sketch it is necessary to make a free body diagram, at the time before and after cutting the cord.
These free body diagrams can be seen in the attached image.
First we perform a sum of forces on the x & y axes before cutting the cord, to be able to find the T tension of the wire. (This analysis can be seen in the attached image).
In this way we get the T-wire tension equation, before cutting.
Now we make another free body diagram, for the moment when the wire is cut (see in the attached diagram).
It is important to clarify that when the cord is cut, the system will no longer be in statically, therefore newton's second law will be used for summation of forces which will be equal to the product of mass by acceleration.
Finally with equations 1 and 2 we can find the K ratio.