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# 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.

### 2 Answers

- 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.

Vertical forces:

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)