Here a cat is running at constant speed which is given as 10 km/h for 5s
So here the average speed is defined as total distance moved in total time interval
so here it is given by
since
here speed of cat is constant so it will remain the same
And hence the average speed and instantaneous speed at any instant for this duration will remain the same
so here answer would be
<em>average speed = 10 km/h</em>
<em>instantaneous speed = 10 km/h</em>
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 = 0
the y-component of the force at A is = 400 N
the couple acting at A is; = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑ = 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 ∑ = 0
Vector force acting at A, =
+
Now, From the image, we have;
a)
∑f = 0
- 600j + 200j = 0i + 0j
+
- 600j + 200j = 0i + 0j
+ (
- 400)j = 0i + 0j
now by equating i- coefficients'
= 0
so, the x-component of the force at A is = 0
also by equating j-coefficient
- 400 = 0
= 400 N
hence, the y-component of the force at A is = 400 N
we also have;
∑ = 0
- ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
- 30 N-m - 228 N-m + 112 Nm = 0
- 146 N-m = 0
= 146 N-m
Therefore, the couple acting at A is; = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑ = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)(
) +
∑ = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑ = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑ = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑ = 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
