Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
Answer:
A force is a push or a pull and it affects our daily lives because without force,people would not be able to open and close stuff or lift up our arms or legs .....or anything, for that matter.
Explanation:
brainly
Answer:
Explanation:
Given:
- mass of vehicle,
- radius of curvature,
- coefficient of friction,
<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>
where:
coefficient of friction
normal reaction force due to weight of the car
velocity of the car
is the maximum velocity at which the vehicle can turn without skidding.
Answer:
1)ammeter
2)ised to check measure of current flow through a circuit
3)o.90 ambere
