Ask question

Ask question

All categories

2 years ago

7

A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force ju

st exceeding 177 N and parallel to the slope is sufficient to cause the block to start moving, find the coefficient of static friction.

1 answer:

stira [4]2 years ago

3 0

Answer:

The coefficient is 0.90

Explanation:

Drawing a diagram makes thing easier, we will assume that the acceleration tends to zero because it start barely moving.

$-F_s+mg*sin(\theta)+F=0\\F_s=57.1kg*9.8m/s^2*sin(28.3)+177N\\F_s=442N\\F_s=\µ*N\\N=m*g*cos(\theta)\\N=57.1*9.8*cos(28.3)=493N\\\\\µ=\frac{442N}{493N}=0.90$

You might be interested in

A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel

omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N .

3 0

3 years ago

A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

T = F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm $= \frac{2}{100} = 0.02m$ for z the perpendicular distance

So Elbow Torque is $T_e= 4000 * 0.02$

$= 80Nm$

To obtain the torque required we substitute 300 N for F and 30cm $=\frac{30}{100} = 0.3 m$

So the Required Torque is $T_R = 300 *0.3$

$=90Nm$

Now since $T_e < T_R$ it mean that the weight lifter would not get past this sticking point

7 0

2 years ago

During strengthening heat treatment, the _______ step traps the material in an unstable crystalline structure. a)-Quenching, b)-

ZanzabumX [31]

Answer: A) Quenching

Hope this helps

4 0

3 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0

3 years ago

he graph shows the distance (x) traveled by an aircraft traveling at constant velocity in corresponding time intervals (t). What

Feliz [49]

Answer:

B linear, 825 meters

Explanation:

5 0

3 years ago