Answer:
50 N
Explanation:
Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:
The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.
The horizontal component of the forces:
F₁ + F₂ = -40N + F₂ = 0
F₂ = 40N
The vertical component of the forces:
F₁ + F₂ - mg = 0 + F₂ - mg = 0
F₂ = mg
If I assume the gravitational constant g = 10 m/s²:
F₂ = (3 kg) * (10 m/s²) = 30N
Adding the horizontal and vertical components of the force F₂:
F₂ = √((40N)² + (30N)²) = 50N
Answer:
12N
Explanation:
Suppose the string mass is negligible, the total mass of the 2 block system is 6 + 9 = 15 kg
So the acceleration of the system when subjected to 30N force is
a = F / M = 30 / 15 = 2 m/s2
So both blocks would have the same acceleration, however, the force acting on the 6kg block would have a magnitude of
f = am = 2 * 6 = 12N
This is the tension in the string between the blocks
<span>Since the torque involves the product of force times lever arm, a small force can exert a greater torque than a larger force if the small force has a large enough lever arm.
With a large force exerts a small torque is a gate, hinged in its vertical line (axis). When pushed from a point near to the hinge, a very large amount is needed to open the gate.
</span><span>
</span>
If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that
<em>n</em> = 38 N
The friction force is proportional to the normal force by a factor of 0.27, so that
<em>f</em> = 0.27 (38 N) ≈ 10.3 N
and so the answer is D.
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
