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erik [133]
3 years ago
5

Joel has a mass of 50kg he is spending his firs day on skis. His friend is trying to push him across a level patch of snow. Joel

friend is pushing with 60N of force but Joel does not move why?
Physics
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

Because of frictional force acting in the opposite direction.

Explanation:

The force of push on the body of Joel is 60 N. The body doesn't move. This clearly means that there must be a force acting in the opposite direction that has a magnitude equal to 60 N so that the net force acting on the body of Joel is 0 and hence the body doesn't move forward.

This opposite force is the frictional force which acts between the surface of snow and Joel. The frictional force always opposes relative motion.

So, when Joel's friend pushes him, he tends move him forward. Therefore, frictional force acts in the opposite direction to oppose the motion. If the frictional force is strong enough to stop the force of push, the body won't move.

This frictional force acting on a stationary body is a variable force and has a maximum value known as limiting friction. When the force of push exceeds the limiting friction, the body just starts to move. The body won't move till the force pf push is less than or equal to limiting friction.

Thus, Joel' friend push is less than or equal to the limiting friction and therefore, Joel is not moving.

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This dude was sitting in the chair. Someone pulls the chair out from under him. He does not move in the direction of the chair.
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Answer:

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ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

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