The frictional force is 218.6 N
Explanation:
The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.
There are two forces acting along this direction:
- The component of the weight parallel to the incline, downward along the plane, of magnitude
where
m = 46 kg is the mass
is the acceleration of gravity
is the angle of the incline
- The (static) frictional force, acting upward, of magnitude
Since the block is in equilibrium, we can write
And substituting, we find the force of friction:
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The acceleration is 2 mph/min
Explanation:
The acceleration of an object is the rate of change of velocity. It is defined as follows:
where
v is the final velocity
u is the initial velocity
t is the time taken for the velocity to change from u to v
For the car in this problem, we have (taking North as positive direction):
u = 25 mph
v = 35 mph
t = 5 min
Substituting into the equation, we find the acceleration:
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Answer:
T = 540 N (to two significant digits)
Explanation:
Let the crate dimension L be from strap attachment to floor contact
Let T be the strap tension
sum moments about the floor contact point to zero
mg[½Lcos25] - Tsin61[Lcos25] + Tcos61[Lsin25] = 0
L is common to all terms, so divides out.
½(71)(9.8)cos25 = T(sin61cos25 - cos61sin25)
T = (71)(9.8)cos25 / (2(sin61cos25 - cos61sin25))
T = 536.428020...