A load of 500N is carried by 200N effort in a simple machine having load distance 3m Calculate effort distance.
1 answer:
Answer:
<h2>2.5 m</h2>Explanation:
Load ( L ) = 500 N
Effort ( E ) = 200 N
Load distance ( LD ) = 3 m
Effort distance ( ED ) = ?
Now, Let's find the Effort distance ( ED )
We know that,
Output work = Input work
i.e L × LD = E × ED
plug the values
multiply the numbers
Swipe the sides of the equation
Divide both sides of the equation by 200
Calculate
Hope this helps..
best regards!!
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Answer:
a = 1 m/s² and
Explanation:
The first two parts can be seen in attachment
We use Newton's second law on each axis
Y axis
Ty - W = 0
Ty = w
X axis
Tx = m a
With trigonometry we find the components of tension
Sin θ = Ty / T
Ty = T sin θ
Cos θ = Tx / T
Tx = T cos θ
We calculate the acceleration with kinematics
Vf = Vo + a t
a = (Vf -Vo) / t
a = (20 -10) / 10
a = 1 m/s²
We substitute in Newton's equations
T Sin θ = mg
T cos θ = ma
We divide the two equations
Tan θ = g / a
θ = tan⁻¹ (g / a)
θ = tan⁻¹ (9.8 / 1)
θ = 84º
We see that in the expression of the angle the mass does not appear therefore you should not change the angle
