To leave Nicholson, you exit via the door to the quad. The door is 40 inches wide and has a mass of 200 kg.
You push the door with a perpendicular force of 52.61N exactly 0.86m away from the hinge. This generates the minimum torque, \tau_1, required to open the door.
If you pushed the door 0.24m away from the hinge, with what force (in N) must you push the door to supply the same torque?
52.6*.86=force*.24
solve for force
Well, isn't this a door-iffic question! Let's see if we can find the solution, shall we?
Seems like you're looking for the force required to supply the same torque when pushing the door at a different distance from the hinge.
To solve this, we'll assume that the torque remains constant between the two situations. We can use the formula torque (τ) = force (F) × distance (r) from the hinge. So, by rearranging the formula, we have force (F) = torque (τ) ÷ distance (r).
In the first situation, τ1 = 52.61 N × 0.86 m (since the force is perpendicular to the door). Let's calculate τ1:
τ1 = 52.61 N × 0.86 m ≈ 45.229 Nm
Now, in the second situation, we can use the same torque (τ1) but with a different distance (r):
τ1 = F × 0.24 m
Now, let's rearrange the formula to solve for F:
F = τ1 ÷ 0.24 m
Plugging in our previous value for τ1:
F = 45.229 Nm ÷ 0.24 m ≈ 188.454 N
So, to generate the same torque when the door is pushed 0.24 m away from the hinge, you'll need to apply a force of approximately 188.454 N. Remember not to clown around too much when dealing with doors!
To find the force required to supply the same torque when pushing the door 0.24m away from the hinge, we can use the formula for torque:
Torque (τ) = Force (F) x Distance (d) x sin(θ)
Since the torque is the same for both cases, we can write:
τ₁ = F₁ x d₁ x sin(θ)
τ₂ = F₂ x d₂ x sin(θ)
where τ₁ and τ₂ are the torques for the first and second cases, F₁ and F₂ are the forces applied, d₁ and d₂ are the distances from the hinge, and θ is the angle between the applied force and the line connecting the hinge and the point of application.
Since the torques are the same, we can set the equations equal to each other:
F₁ x d₁ x sin(θ) = F₂ x d₂ x sin(θ)
We can rearrange the equation to solve for F₂:
F₂ = (F₁ x d₁) / d₂
Now we can substitute the given values into the equation:
F₁ = 52.61 N (given)
d₁ = 0.86 m (given)
d₂ = 0.24 m (given)
F₂ = (52.61 N x 0.86 m) / 0.24 m
F₂ = 188.6825 N
Therefore, the force required to supply the same torque when pushing the door 0.24m away from the hinge is approximately 188.6825 N.
To find the force required to generate the same torque when pushing the door at a different distance from the hinge, we can use the concept of torque.
Torque is calculated as the product of the force applied and the perpendicular distance from the point of rotation (in this case, the hinge). Mathematically, torque (τ) is given by:
τ = F * r
where τ is torque, F is the force applied, and r is the perpendicular distance from the point of rotation (hinge).
Given:
- τ1 = minimum torque required to open the door (when pushed at a distance of 0.86m from the hinge)
- F1 = force applied to generate τ1
- r1 = distance from the hinge when pushing with F1 (0.86m)
- r2 = distance from the hinge when we need to find the force (0.24m)
We can set up the equation using the torque formula:
τ1 = F1 * r1
To find the force required when pushing at a distance of 0.24m, we rearrange the equation to solve for F2:
F2 = τ1 / r2
Now, substitute the given values:
F2 = τ1 / r2 = (F1 * r1) / r2
Substitute the known values:
F1 = 52.61N
r1 = 0.86m
τ1 is the minimum torque required to open the door.
Now we can calculate F2:
F2 = (52.61N * 0.86m) / 0.24m
Simplifying this equation:
F2 = 188.3283N
Therefore, to supply the same torque when pushing the door at a distance of 0.24m from the hinge, the force required is approximately 188.33N.