Help with physics question!!?
A 820 N crate is being pulled across a level floor by a force F of 400 N at an angle of 33° above the horizontal. The coefficient of kinetic friction between the crate and the floor is 0.25. Find the magnitude of the acceleration of the crate.
- 1 decade agoFavorite Answer
First find Fn
Sum Forces = Fg - Fn - Fay = 0
Fay = 217.86 N
Fn = 602.14 N
Then Find Ffr
Ffr = mu * Fn = .25 * 602.14 = 150.54
Then find a
Sum Fx = Fax - Ffr = ma
335.47 N - 150.54 N = 83.67 kg * a
a = 2.21 m/s^2
- RickBLv 71 decade ago
Figure out the net force acting on the crate. Then use the famous equation "Fnet=ma" to solve for the acceleration.
In this problem, it's easiest to look at the vertical forces and horizontal forces separately.
a. Weight = mg = 820N (down)
b. Normal force from floor = F_n (up) [note we don't know how much that is yet.]
c. Upward component of the force due to the rope (I guess it's a rope). This is (400N)(sin33)
Since there is no acceleration in the vertical direction, this must mean the vertical forces are balanced (upward forces = downward forces). That is:
(1): F_n + (400N)(sin33) = 820N
Horizontal forces (assume crate is being pulled toward the right):
a. Horizontal component due to rope (pulling right). This is (400N)(cos33)
b. Friction (pulling left). this is (F_n)(μ)
The net force is the difference between the right-pulling force and the left-pulling force:
(2): Fnet = (400N)(cos33) − (F_n)(μ)
Now combine (1) with (2) to eliminate the variable F_n:
(3): Fnet = (400N)(cos33) − (820N−400N(sin33))(μ)
Finally, divide Fnet by m, to get the acceleration. (Note that m = weight/g = 820N/g)
a = Fnet/m = Fnet(g/820N)
= ((400N)(cos33) − (820N−400N(sin33))(μ))(g/820N)