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solong [7]
3 years ago
15

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to

an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256m/s
Physics
1 answer:
Salsk061 [2.6K]3 years ago
7 0

Explanation:

1.2 \mathrm{N} ; 2 \mathrm{N}

2.200 \mathrm{N} ; 200 \mathrm{N}

4.2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}

5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}

6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}

8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of 256 \mathrm{m} / \mathrm{s}

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

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The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a)
Marysya12 [62]

This question is incomplete, the complete question is;

The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.

(a) Determine the forces and and the couple

(b) Determine the sum of the moments about the right end of the beam.

(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?

Answer:

a)

the x-component of the force at A is A_{x} = 0

the y-component of the force at A is A_{y}  = 400 N

the couple acting at A is; M_{A} = 146 N-m

b)

the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

the equivalent force acting at the left end is; F = -400J ( N)

the couple acting at the left end is; M = - 146 N-m

Explanation:

Given that;

The sum of the forces acting on the beam is zero ∑f = 0

Sum of the moments about the left end of the beam is also zero ∑M_{L} = 0

Vector force acting at A, F_{A} = A_{x}i + A_{y}j

Now, From the image, we have;

a)

∑f = 0

F_{A} - 600j + 200j = 0i + 0j

A_{x}i + A_{y}j - 600j + 200j = 0i + 0j

A_{x}i + (A_{y} - 400)j = 0i + 0j

now by equating i- coefficients'

A_{x} = 0

so, the x-component of the force at A is A_{x} = 0

also by equating j-coefficient

A_{y} - 400 = 0

A_{y}  = 400 N

hence, the y-component of the force at A is A_{y}  = 400 N

we also have;

∑M_{L} = 0

M_{A}  - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0

M_{A} - 30 N-m - 228 N-m + 112 Nm = 0

M_{A} - 146 N-m = 0

M_{A} = 146 N-m

Therefore, the couple acting at A is; M_{A} = 146 N-m

b)

The sum of the moments about right end of the beam is;

∑M_{R} = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(A_{y} ) + M_{A}

∑M_{R} = (108  N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m

∑M_{R} = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m

∑M_{R}  = 0

Therefore, the sum of the momentum about the right end of the beam is;  ∑M_{R}  = 0

c)

The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.

The equivalent force at the left end will be;

F = -600j + 200j (N)

F = -400J ( N)

Therefore, the equivalent force acting at the left end is; F = -400J ( N)

Also couple acting at the left end

M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)

M = -(30 N-m) + (112 N-m) - ( 228 N-m))

M = 112 N-m - 258 N-m

M = - 146 N-m

Therefore, the couple acting at the left end is; M = - 146 N-m

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