# Search results

1. ### Can someone ! Equilibrium/friction

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.
2. ### Can someone ! Equilibrium/friction

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.
3. ### Can someone ! Equilibrium/friction

[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...
4. ### Equilibrium/friction problem

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.
5. ### Equilibrium/friction problem

I don't understand.
6. ### Equilibrium/friction problem

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
7. ### Equilibrium/friction problem

[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...
8. ### Laws of Motion problem

nevermind I got it.

ax is wrong?
10. ### Laws of Motion problem

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...
11. ### Relative velocity of two cars

thanks, the textbooks never have examples like the ones you have to do in the homework, it makes it very difficult.
12. ### Relative velocity of two cars

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
13. ### Relative velocity of two cars

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
14. ### Relative velocity of two cars

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...
15. ### Yet another projectile motion

sin(2theta)=.00492, can't remember how to get theta from here.
16. ### Yet another projectile motion

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)
17. ### Yet another projectile motion

from v0 in a one dimensional equation.
18. ### Yet another projectile motion

what do I do with the t variable? Is it the time is takes to go 91.4m at 427m/s? .214s.
19. ### Yet another projectile motion

91.4m=vx*cos(theta)t, 0m=vy*sin(theta)t + 1/2(-9.80 m/s^2)t^2
20. ### Yet another projectile motion

is the v0 the same throughout the problem, or is it different for v0x, and voy?
21. ### Yet another projectile motion

x-x0=delta x, y-y0=delta y, delta x=91.4m, delta y=0
22. ### Yet another projectile motion

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
24. ### Projectile motion rocket problem

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...
25. ### Acceleration to displacement

nevermind, I got it now!
26. ### Acceleration to displacement

x=(0.300m/s^2 *t^2)/2 ?
27. ### Acceleration to displacement

[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
29. ### Acceleration problem

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

a2=2.5v-v/t
31. ### Acceleration problem

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

[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...