In the figure, a block weighing 22.0 N is held at rest against a vertical wall b

ID: 1883393 • Letter: I

Question

In the figure, a block weighing 22.0 N is held at rest against a vertical wall by a horizontal forceof magnitude 60 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second forceis applied to the block and directed parallel to the wall with these magnitudes and directions: (a)33 N, up, (b)12 N, up, (c)49 N, up, (d)63 N, up, (e)9.2 N, down, and (f)19 N, down. In each experiment, what is the frictional force on the block, including sign? Take the direction up the wall as positive, and down the wall as negative. Next, calculate the acceleration, including sign, of the block irn each case. Note that acceleration is zero if the block does not move (g) What is the acceleration in (a)? (h) What is the acceleration in (b)? (i) What is the acceleration in (c)? (i) What is the acceleration in (d)? (k) What is the acceleration in (e)? (I) What is the acceleration in (f)?

Explanation / Answer

GIven :

N = 60 N (force by block on wall) [ in right direction]

FN = 60 N (normal reaction by wall on block) [in left direction]

us = 0.55 (coefficient of static friction)

fs = us * N = 33 N (maximum value of static friction)

uk = 0.38 (coefficient of kinetic friction)

fk = uk * N = 22.8 N (value of kinetic friction, once body starts moving)

m*g = W = 22 N (weight of the block) [in downward direction] ; m = 2.2 kg (assuming g = 10 m/s2)

P = 33 N (applied force) [ in up direction]

So, by equating forces along vertical:

So, Fnet along vertical : P - fs - W

On solving, Fnet = 33 - 33 - 22 = -22 < 0 (Assuming maximum static friction)

Here, we can see that weight of body < F applied, so now whether the body will move or not depends on friction.

Here, by taking friction as maximum we are getting net force = -22 N, which implies that the body won't move.

So, the value of frictional force acting on the block:

Here, since fs can vary, so make fs such that Fnet becomes 0. So if fs = 11 N, then we can see that Fnet = 0. So, value of frictional force acting on the block is : 11 N (in down direction) = -11N

P = 12 N (applied force) [ in up direction]

So, by equating forces along vertical:

So, Fnet along vertical : P - fs - W

On solving, Fnet = 12 + 33 - 22 = 23 > 0 (Assuming maximum static friction)

Here, the acting friction will be static and that also in the also in upward direction. Since F applied < Weight. The difference will be handled by static friction to keep the body at rest.

So, the value of frictional force acting on the block:

Here, we need static friction such that it makes net force = 0, So such value of static friction : 10 N (in up direction) = +10 N

P = 49 N (applied force) [ in up direction]

So, by equating forces along vertical:

So, Fnet along vertical : P - fs - W

On solving, Fnet = 49 - 33 - 22 = -6 < 0 (Assuming maximum static friction)

Here, we can see that weight of body < F applied, so now whether the body will move or not depends on friction.

Here, by taking friction as maximum we are getting net force = -6 N, which implies that the body won't move.

So, the value of frictional force acting on the block:

Here, since fs can vary, so make fs such that Fnet becomes 0. So if fs = 27 N, then we can see that Fnet = 0. So, value of frictional force acting on the block is : 27 N (in down direction) = -27 N

P = 63 N (applied force) [ in up direction]

So, by equating forces along vertical:

So, Fnet along vertical : P - fs - W

On solving, Fnet = 63 - 33 - 22 = 7 > 0 (Assuming maximum static friction)

This implies, that body will move from its location.

So, the value of frictional force acting on the block:

Since, the body has start moving, kinetic friction will be acting on the block. So, value of frictional force : 22.8 N (in down direction) = -22.8 N

P = 9.2 N (applied force) [ in down direction]

So, by equating forces along vertical:

So, Fnet along vertical : -P + fs - W

On solving, Fnet = -9.2 + 33 - 22 = 1.8 > 0 (Assuming maximum static friction)

So, here the body still has to overcome 1.8 N of extra force to be able to move.

Since, body is in rest, static friction is being applied on the body.

So, the value of frictional force acting on the block:

Here, since fs can vary, so make fs such that Fnet becomes 0. So if fs = 31.2 N, then we can see that Fnet = 0. So, value of frictional force acting on the block is : 31.2 N ( in up direction) = +31.2 N

P = 19 N (applied force) [ in down direction]

So, by equating forces along vertical:

So, Fnet along vertical : -P + fs - W

On solving, Fnet = -19 + 33 - 22 = -8 < 0 (Assuming maximum static friction)

Here, it implies that the maximum static friction is not able to stop the body and is short by 8 N.

So, here body will be in motion, so kinetic friction will be retarding its motion.

So, the value of frictional force acting on the block: Since, it is in motion. Kinetic friction acts on it. So, value of frictional force acting on the mass : 22.8 N (in up direction) = + 22.8 N

Since, static friction is acting on the block implying the block is in rest, acceleration of block a = 0 m/s2

Since, kinetic friction is acting on the block implying motion:

So, by equating forces along vertical:

m*a = -m*g + F applied - fk

2.2 * a = -22 + 63 - 22.8

2.2 * a = 18.2 N

on solving,

a = +8.27 m/s2

Since, static friction is acting on the block implying the block is in rest, acceleration of block a = 0 m/s2

Since, kinetic friction is acting on the block implying motion:

So, by equating forces along vertical:

m*a = -m*g - F applied + fk

2.2 * a = - 22 - 19 + 22.8

2.2 * a = -18.2 N

on solving,

a = -8.27 m/s2

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