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a)Fs=\musFn Fn=F(cos40) Fs=(.560)(F(cos40)) Fs-mg+F(sin40)=0 F(0.560*cos(40)+sin(40))=88.9N F=82.9N b)-Fs-mg+F(sin(40))=0 F(sin(40)-(.560*cos(40))=88.9N F=415.4N answer for b doesn't seem right.

a) Fs acts up b)Fs acts down The problem is I'm not sure how to get Fs from the normal force in the x direction from the wall.

[bQuestion Details: I am having trouble answering this question. The drawing shows a block against a vertical wall to its right. The force F is pointed at the lower left hand corner of the block at 40.0 deg. left of -y-axis The weight of the block in the drawing is 88.9N. The...

No, I think I am having trouble because I am not sure where to place the applied force vector on my free body diagram. I have fs pointing +y, mg -y, Fn=-x, F 40 deg. in the (-x,-y)quadrant.

I don't understand.

a)88.9N*tan(40)=74.6N=Fn Fy=88.9N Fs=(0.560)*(74.6N)=41.8N 88.9N-41.8N=47.1N \sqrt{47.1^2+74.6^2}=88.2N F=88.2N

[b]1. There is a block parallel and touching a vertical wall, a force F is applied at a 40 deg. angle from -x,-y. the weight of the block is 88.9N. The coefficient of static friction between the block and the wall is 0.560. a) what is the minimum force F required to prevent the block from...

nevermind I got it.

ax is wrong?

Homework Statement Only two forces act on an object (mass=3.00kg). F1=60.0N 45.0deg above the +x axis, F2=40.0N on the +x axis. Find the magnitude and direction (relative to the x axis) of the acceleration of the object. Homework Equations F=ma, The Attempt at a Solution...

thanks, the textbooks never have examples like the ones you have to do in the homework, it makes it very difficult.

Oh, it is not clear from my text. So, (24.4m/s-18.6m/s=5.8m/s) 186m=(5.8m/s)t t=32.1s

for relative velocity, velocity of A with respect to B is Va+Vb=Vab (24.4m/s+18.6m/s=43.0m/s). 186m=(43.0 m/s)*t t=4.33s

Homework Statement Two cars, A and B, are traveling in the same direction, although car A is 186m behind car B. The speed of A is 24.4m/s, and the speed of B is 18.6m/s. how much time does it take for A to catch B? Homework Equations Va+Vb=Vab, x=1/2(V0x+Vx)t The Attempt at a...

sin(2theta)=.00492, can't remember how to get theta from here.

0m=(427m/s)*sin(theta)*((91.4m)/((427m/s)*cos(theta))+(1/2)*(-9.80m/s^2)*((91.4m)/((427m/s)*cos(theta))^2 (0.225/cos^2(theta))=(91.4*sin(theta))/cos(theta)

from v0 in a one dimensional equation.

what do I do with the t variable? Is it the time is takes to go 91.4m at 427m/s? .214s.

91.4m=vx*cos(theta)t, 0m=vy*sin(theta)t + 1/2(-9.80 m/s^2)t^2

is the v0 the same throughout the problem, or is it different for v0x, and voy?

x-x0=delta x, y-y0=delta y, delta x=91.4m, delta y=0

Homework Statement A rifle has been sighted in for a 91.4m target. If the muzzle speed of the bullet is v0=427 m/s, what are the two possible angles theta1 and theta2 betweent the rifle barrel and the horizontal such that the bullet will hit the target? One of the angles is so large that it is...

thanks

Homework Statement A rocket is fired at at speed fo 75.0 m/s from ground level, at an angle of 60.0 deg. above the horizontal. The rocket is fired toward an 11.0 m high wall, shich is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after...

nevermind, I got it now!

x=(0.300m/s^2 *t^2)/2 ?

[b]1. two knights on horseback start from rest 88.0 m apart and ride directly toward each other to do battle. Sir George's acceleation has a magnitude of 0.300 m/s^2, Sir Alfred's has a magnitude of 0.200 m/s^2. Relative to Sir George's starting point, where do the knights collide? [b]2...

thanks

comes out to 4.5 m/s^2 still, right?

a2=2.5v-v/t

velocity at end of stage 1: v=(3.0m/s^2*t)?

[b]1. An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is 3.0m/s^2. The magnitude of the car's velocity at the end of stage 2 is 2.5 times...