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mrs_skeptik [129]

3 years ago

11

Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing ind

icate. If the coefficient of static friction between a crutch and the ground is 0.913, determine the largest angle MAX that the crutch can have just before it begins to slip on the floor.

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Answer: 43 degrees

Explanation:

The force of the crutch can be broken into components. The horizontal component of F is the static friction force keeping the crutch from sliding. The vertical component is opposing the weight and is the Normal force. Using the orientation of the angle q, we have the following

fs = Fx = F sin (angle (tita))

N = Fy = F cos (angle(tita))

Maximum angle implies maximum static friction

Therefore,

fsmax = UsN = Us x F cos(angle tita)

Where U = miyu

F sin(angle tita) = Us x Fcos (angle tita)

Sin (angle tita) / cos (angle tita) = Us

Therefore, tan (angle tita) = Us

Angle tita = tan^-1(Us) = tan^-1 (0.931) = 42.95 degrees = 43 degrees

There for Angle tita Max = 43 degrees

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zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

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$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

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